In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of a single ball from a collection of identical but different coloured balls. e.g. An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. If one ball is drawn at random from the urn. Find the probability that of getting a red ball.

Problems based on the Draw of a Single Ball:

Example – 01:

An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. If one ball is drawn at random from the urn. Find the probability that

Solution:

Total = 9 + 7 + 4 = 20

There are 20 balls in the urn

one ball out of 20 can be drawn by ²⁰C₁ ways

Hence n(S) = ²⁰C₁ = 20

a) a red ball

Let A be the event of getting a red ball

There are 9 red balls in the urn

1 red ball out of 9 red balls can be drawn by ⁹C₁ ways

∴ n(A) = ⁹C₁  = 9

By the definition P(A) = n(A)/n(S) =9/20

Therefore the probability of getting a red ball is 9/20.

b) a white ball

Let B be the event of getting a white ball

There are 7 white balls in the urn

1 white ball out of 7 white balls can be drawn by ⁷C₁ ways

∴ n(B) = ⁷C₁  = 7

By the definition P(B) = n(B)/n(S) =7/20

Therefore the probability of getting a white ball is 7/20.

c) a black ball

Let C be the event of getting a black ball

There are 4 black balls in the urn

1 black ball out of 4 black balls can be drawn by ⁴C₁ ways

∴ n(C) = ⁴C₁  = 4

By the definition P(C) = n(C)/n(S) = 4/20 = 1/5

Therefore the probability of getting a black ball is 1/5.

d) not a red ball

Let D be the event of getting not a red ball

There are 7 + 4 = 11 non-red balls

1 non-red ball out of 11 non-red balls can be drawn by ¹¹C₁ ways

∴ n(D) = ¹¹C₁  = 11

By the definition P(D) = n(D)/n(S) = 11/20

Therefore the probability of getting not red ball is 11/20

e) not a white ball

Let E be the event of getting not a white ball

There are 9 + 4 = 13  non-white balls

1 non-white ball out of 13 non-white balls can be drawn by ¹³C₁ ways

∴ n(E) = ¹³C₁  = 13

By the definition P(E) = n(E)/n(S) = 13/20

Therefore the probability of getting not white ball is 13/20

f) not a black ball

Let F be the event of getting not a black ball

There are 9 + 7 = 16 non-black balls

1 non-black ball out of 16 non-black balls can be drawn by ¹⁶C₁ ways

∴ n(F) = ¹⁶C₁  = 16

By the definition P(F) = n(F)/n(S) = 16/20 = 4/5

Therefore the probability of getting not a black ball is 4/5.

g) a red ball or a black ball

Let G be the event of getting not a red ball

There are 9 + 4 = 13 red or black balls

1 red or a  black ball out of 13 can be drawn by ¹³C₁ ways

∴ n(G) = ¹³C₁  = 13

By the definition P(G) = n(G)/n(S) = 13/20

Therefore the probability of getting a red ball or a black ball is 13/20.

h) a red ball or a white ball

Let H be the event of getting a red ball or a white ball

There are 9 + 7 = 16 red or white balls

1 red or a  white ball out of 16 can be drawn by ¹⁶C₁ ways

∴ n(H) = ¹⁶C₁  = 16

By the definition P(H) = n(H)/n(S) = 16/20 = 4/5

Therefore the probability of getting a red ball or a white ball is 4/5.

i) a black ball or a white ball

Let J be the event of getting a black ball or a white ball

There are 4 + 7 = 11 black or white balls

1 black or a  white ball out of 11 can be drawn by ¹¹C₁ ways

∴ n(J) = ¹¹C₁  = 11

By the definition P(J) = n(J)/n(S) = 11/20

Therefore the probability of getting a black ball or a white ball is 11/20.

Example – 02:

An urn contains 9 red, 7 white, and 4 black balls.  All balls are identical. Two balls are drawn at random from the urn. Find the probability that

Solution:

Total = 9 + 7 + 4 = 20

There are 20 balls in the urn

Two balls out of 20 can be drawn by ²⁰C₂ ways

Hence n(S) = ²⁰C₂ = 10 x 19

a) both red balls

Let A be the event of getting both red balls

There are 9 red balls in the urn

2 red balls out of 9 red balls can be drawn by ⁹C₂ ways

∴ n(A) = ⁹C₂  = 9 x 4

By the definition P(A) = n(A)/n(S) = (9 x 4)/(10 x 19) = 18/95

Therefore the probability of getting both red balls is 18/95

b) no red ball

Let B be the event of getting no red ball

There are 7 + 4 = 11 non-red balls in the urn

2 red balls out of 11 non-red balls can be drawn by ¹¹C₂ ways

∴ n(B) = ¹¹C₂  = 11 x 5

By the definition P(B) = n(B)/n(S) = (11 x 5)/(10 x 19) = 11/38

Therefore the probability of getting no red ball is 11/76

c) atleast one red ball

Let C be the event of getting atleast one red ball

Hence C is the event of getting no red ball

There are 7 + 4 = 11 non-red balls in the urn

2 non-red balls out of 11 non-red balls can be drawn by ¹¹C₂ ways

∴ n(C) = ¹¹C₂  = 11 x 5

By the definition P(C) = n(C)/n(S) = (11 x 5)/(10 x 19) = 11/38

Now, P(C) = 1 – P(C) = 1 – 11/38 = 27/38

Therefore the probability of getting atleast one red ball is 27/38

d) exactly one red ball

Let D be the event of getting exactly one red ball i.e. one red and 1 non red ball

There are 9 red and 7 + 4 = 11 non-red balls in the urn

2 red balls out of 11 non-red balls can be drawn by ¹¹C₂ ways

∴ n(D) = ⁹C₁  x  ¹¹C₁  = 9  x 11

By the definition P(D) = n(D)/n(S) = (9 x 11)/(10 x 19) = 99/190

Therefore the probability of getting exactly one red ball is 99/190.

e) at most one red ball

Let E be the event of getting at most one red ball There are two possibilities

Case – 1: Getting no red ball and two non-red balls or

Case – 2: Getting 1 red ball and 1 non-red ball

There are 9 red and 7 + 4 = 11 non-red balls in the urn

∴ n(E) = ⁹C₀  x  ¹¹C₂ +  ⁹C₁  x  ¹¹C₁ = 1 x 11 x 5 + 9 x 11 = 55 + 99 = 154

By the definition P(E) = n(E)/n(S) = 154/(10 x 19) = 77/95

Therefore the probability of getting at most one red ball is 77/95.

f) one is red and other is white

Let F be the event of getting one red and one white ball

There are 9 red and 7 white balls in the urn

∴ n(F) = ⁹C₁  x  ⁷C₁ = 9 x 7

By the definition P(F) = n(F)/n(S) = (9 x 7)/(10 x 19) = 63/190

Therefore the probability of getting one red and other white ball is 63/190

g) one is red and other is black

Let G be the event of getting one red and one black ball

There are 9 red and 4 black balls in the urn

∴ n(G) = ⁹C₁  x  ⁴C₁ = 9 x 4

By the definition P(G) = n(G)/n(S) = (9 x 4)/(10 x 19) = 18/95

Therefore the probability of getting one red and other black ball is 18/95

h) one is white and the other is black

Let H be the event of getting one white and one black ball

There are 7 white and 4 black balls in the urn

∴ n(H) = ⁷C₁  x  ⁴C₁ = 7 x 4

By the definition P(H) = n(H)/n(S) = (7 x 4)/(10 x 19) = 14/95

Therefore the probability of getting one white and other black ball is 14/95.

i) both are of same colour

Let J be the event of getting balls of same colour.

i.e. both are red or both are white or both are black

There are 9 red, 7 white, and 4 black balls in the urn

∴ n(J) = ⁹C₂  +  ⁷C₂ +  ⁴C₂ = 9 x 4 + 7 x 3 + 2 x 3 = 36 +21 + 6 = 63

By the definition P(J) = n(J)/n(S) = 63/(10 x 19) = 63/190

Therefore the probability of getting both balls of same colour is 63/190

j) both are of not of the same colour

Let K be the event of getting balls not of the same colour.

Hence K is the event of getting the balls of same colour

i.e. both are red or both are white or both are black

There are 9 red, 7 white, and 4 black balls in the urn

∴ n(K) = ⁹C₂  +  ⁷C₂ +  ⁴C₂ = 9 x 4 + 7 x 3 + 2 x 3 = 36 +21 + 6 = 63

By the definition P(K) = n(K)/n(S) = 63/(10 x 19) = 63/190

Now, P(K) = 1 – P(K) = 1 – 63/190 = 127/190

Therefore the probability of getting both balls of not of the same colour is 127/190.

k) both are red or both are black

Let L be the event of getting both red or both black balls.

There are 9 red and 4 black balls in the urn

∴ n(L) = ⁹C₂ +  ⁴C₂ = 9 x 4 + 2 x 3 = 36 + 6 = 42

By the definition P(L) = n(L)/n(S) = 42/(10 x 19) = 21/95

Therefore the probability of getting both red or both black balls is 21/95.

In the next article, we shall study some basic problems of probability based on the drawing of two or more balls from a collection of identical coloured balls.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Selection of Balls appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of four or more playing cards. For e.g. Four cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting all the cards of the same suite.

Problems based on the Draw of 4 Playing Cards:

Example – 01:

Four cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

Four cards out of 52 can be drawn by ⁵²C₄ ways

Hence n(S) = ⁵²C₄

a) all are heart cards

Let A be the event of getting all heart cards

There are 13 heart cards in a pack

four heart cards out of 13 heart cards can be drawn by ¹³C₄ ways

∴ n(A) = ¹³C₄

By the definition P(A) = n(A)/n(S) = (¹³C₄ )/(⁵²C₄)

Therefore the probability of getting all heart cards is 

(¹³C₄ )/(⁵²C₄)

b) all the cards are of the same suite

Let B be the event of getting all the cards of the same suite

There are 4 suites in a pack

There are 13 cards in each suite

four cards of the same suite out of 13 cards of same suite can be drawn by ¹³C₄ ways

∴ n(B) = 4 x ¹³C₄

By the definition P(B) = n(B)/n(S) = (4 x ¹³C₄ )/(⁵²C₄)

Therefore the probability of getting all the cards of the same suite is 

(4 x ¹³C₄ )/(⁵²C₄)

c) all the cards of the same colour

Let C be the event of getting all the cards of the same colour.

There are 26 red and 26 black cards in a pack

Thus the selection is all red or all black.

∴ n(C) = ²⁶C₄ + ²⁶C₄ = 2(²⁶C₄)

By the definition P(C) = n(C)/n(S) = 2(²⁶C₄)/(⁵²C₄)

Therefore the probability of getting all the cards of the same colour is

2(²⁶C₄)/(⁵²C₄)

d) all the face cards

Let D be the event of getting all the face cards.

There are 12 face cards in a pack

∴ n(D) = ¹²C₄

By the definition P(D) = n(D)/n(S) = ¹²C₄/(⁵²C₄)

Therefore the probability of getting all face cards is

¹²C₄/(⁵²C₄)

e) all the cards are of the same number (denomination)

Let E be the event of getting all the cards of the same number

there are 4 cards of the same denomination in a pack and 1 in each suite.

There are such 13 sets

four  cards of the same number out of 4 cards can be drawn by ⁴C₄ ways

∴ n(E) = 13 x ⁴C₄  = 13 x 1 = 13

By the definition P(E) = n(E)/n(S) = (13)/(⁵²C₄)

Therefore the probability of getting all the cards of the same suite is 

(13)/(⁵²C₄)

f) Two red cards and two black cards

Let F be the event of getting two red cards and two black cards

There are 26 red and 26 black cards in a pack

∴ n(F) = (²⁶C₂) x (²⁶C₂) = (²⁶C₂)²

By the definition P(F) = n(F)/n(S) = (²⁶C₂)²/(⁵²C₄)

Therefore the probability of getting two red cards and two black cards is

(²⁶C₂)²/(⁵²C₄)

g) all honours of the same suite

Let G be the event of getting honours of the same suite

There are 4 honours (ace, king, queen, and jack) in a suite.

There are four suites

∴ n(G) = (⁴C₄) + (⁴C₄) + (⁴C₄) + (⁴C₄) = 1 + 1 + 1 + 1 = 4

By the definition P(G) = n(G)/n(S) = 4/(⁵²C₄)

Therefore the probability of getting all honours of the same suite is

4/(⁵²C₄)

h) atleast one heart

Let H be the event of getting at least one heart

Thus H’ is an event of getting no heart

Thus there are 39 non-heart cards in a pack

four non-heart cards out of 39 non-heart cards can be drawn by ³⁹C₄ ways

∴ n(H’) = ³⁹C₄

By the definition P(H’) = n(H’)/n(S) = ³⁹C₄/⁵²C₄

Now P(H) = 1 – P(H’) = 1 – (³⁹C₄/⁵²C₄)

Therefore the probability of getting at least one heart is

(1 – (³⁹C₄/⁵²C₄))

i) 3 kings and 1 jack

Let J be the event of getting 3 kings and 1 jack

There are 4 kings and 4 jacks in a pack

∴ n(J) = (⁴C₃ x ⁴C₁)

By the definition P(J) = n(J)/n(S) = (⁴C₃ x ⁴C₁)/(⁵²C₄)

Therefore the probability of getting 3 kings and one jack is

(⁴C₃ x ⁴C₁)/(⁵²C₄)

j) all clubs and one of them is a jack

Let K be the event of getting all clubs and one of them is a jack

There are 12 club  cards + 1 club jack i.e. total 13 club cards

∴ n(K) = (¹²C₃ x ¹C₁) = ¹²C₃

By the definition P(K) = n(K)/n(S) = (¹²C₃)/(⁵²C₄)

Therefore the probability of getting all clubs and one of them is a jack is

(¹²C₃)/(⁵²C₄)

k) 3 diamonds and 1 spade

Let L be the event of getting 3 diamonds and 1 spade

There are 13 diamond cards and 13 spade cards in a pack

∴ n(L) = (¹³C₃ x ¹³C₁)

By the definition P(L) = n(L)/n(S) = (¹³C₃ x ¹³C₁)/(⁵²C₄)

Therefore the probability of getting 3 diamonds and 1 spade is 

(¹³C₃ x ¹³C₁)/(⁵²C₄)

Example – 02:

In a shuffling a pack of 52 cards, four are accidentally dropped, find the probability that the missing cards should be one from each suite.

Solution:

There are 52 cards in a pack.

Four cards out of 52 can be drawn by ⁵²C₄ ways

Hence n(S) = ⁵²C₄

Let A be the event of getting one card from each suite

There are 13 cards in each suite.

∴ n(A) = (¹³C₁) x (¹³C₁) x (¹³C₁) x (¹³C₁)x = (¹³C₁)⁴

By the definition P(A) = n(A)/n(S) = (¹³C₁)⁴/(⁵²C₄)

Therefore the probability of getting one card from each suite is

(¹³C₁)⁴/(⁵²C₄)

Example – 03:

Five cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

Five cards out of 52 can be drawn by ⁵²C₅ ways

Hence n(S) = ⁵²C₅

a) just one ace

Let A be the event of getting just one ace

There are 4 aces and 48 non-aces in a pack

∴ n(A) = ⁴C₁ x ⁴⁸C₄  = 4 x ⁴⁸C₄

By the definition P(A) = n(A)/n(S) = (4 x ⁴⁸C₄ )/(⁵²C₅)

Therefore the probability of getting all heart cards is

(4 x ⁴⁸C₄ )/(⁵²C₅)

b) atleast one ace

Let B be the event of getting atleast one ace

Hence B’ is the event of getting no ace

There are 4 aces and 48 non-aces in a pack

∴ n(B’) =  ⁴⁸C₅

By the definition P(B’) = n(B’)/n(S) = (⁴⁸C₅ )/(⁵²C₅)

Now P(B) = 1 – P(B’) = 1 –  (⁴⁸C₅ )/(⁵²C₅)

Therefore the probability of getting atleast one ace is

(1 –  (⁴⁸C₅ )/(⁵²C₅))

c) all cards are of hearts

Let C be the event of getting all hearts

There are 13 heart cards in a pack

∴ n(C) =  ¹³C₅

By the definition P(C) = n(C)/n(S) = (¹³C₅ )/(⁵²C₅)

Therefore the probability of getting all hearts is

(¹³C₅ )/(⁵²C₅)

Example – 04:

What is the probability of getting 9 cards of the same suite in one hand at a game of bridge?

Solution:

There are 52 cards in a pack.

In a game of bridge, each player gets 13 cards in a hand.

13 cards out of 52 can be drawn by ⁵²C₁₃ ways

Hence n(S) = ⁵²C₁₃

Let B be the event of getting 9 cards of the same suite in one hand

There are 4 suites, thus the suite can be selected by ⁴C₁ ways = 4 ways

Now, in hand, there are 9 cards of same suite and 4 cards of other suites.

∴ n(B) = (⁴C₁) x (¹³C₉) x (³⁹C₄) =  4 x (¹³C₉) x (³⁹C₄)

By the definition P(B) = n(B)/n(S) = (4 x (¹³C₉) x (³⁹C₄) )/(⁵²C₄)

Therefore the probability of getting 9 cards of the same suite in one hand is

(4 x (¹³C₉) x (³⁹C₄) )/(⁵²C₄)

Example – 05:

What is the probability of getting 9 cards of the spade in one hand at a game of bridge?

Solution:

There are 52 cards in a pack.

In a game of bridge, each player gets 13 cards in a hand.

13 cards out of 52 can be drawn by ⁵²C₁₃ ways

Hence n(S) = ⁵²C₁₃

Let C be the event of getting 9 cards of spade in one hand

Now, in hand, there are 9 cards of spade and 4 cards are non-spade.

∴ n(C) = (¹³C₉) x (³⁹C₄) =  (¹³C₉) x (³⁹C₄)

By the definition P(C) = n(C)/n(S) = (¹³C₉ x ³⁹C₄)/(⁵²C₄)

Therefore the probability of getting 9 cards of spade in one hand is

(¹³C₉ x ³⁹C₄)/(⁵²C₄)

Example – 06:

In a hand at whist, what is the probability that four kings are held by a specified player?

Solution:

There are 52 cards in a pack.

In a game, each player gets 13 cards in a hand.

13 cards out of 52 can be drawn by ⁵²C₁₃ ways

Hence n(S) = ⁵²C₁₃

Let D be the event that four kings are held by a specified player

A particular player can be chosen by 1 way

Now, in hand, there are  4 kings and 48 non-king cards

∴ n(D) = (1) x (⁴C₄) x (⁴⁸C₉) =  ⁴⁸C₉

By the definition P(D) = n(D)/n(S) = (⁴⁸C₉ )/(⁵²C₄)

Therefore the probability of that four kings are held by a specified player is  (⁴⁸C₉ )/(⁵²C₄)

Example – 07:

The face cards are removed from a full pack. Out of 40 remaining cards, 4 are drawn at random. find the probability that the selection contains one card from each suite.

Solution:

There are 52 cards in a pack.

There are 12 face cards which are removed. Thus 40 cards remain

Four cards out of 40 can be drawn by ⁴⁰C₄ ways

Hence n(S) = ⁴⁰C₄

Let E be the event of getting one card from each suite

There are 10 cards in each suite.

∴ n(E) = (¹⁰C₁) x (¹⁰C₁) x (¹⁰C₁) x (¹⁰C₁)x = (¹⁰C₁)⁴

By the definition P(E) = n(E)/n(S) = (¹⁰C₁)⁴/(⁴⁰C₄)

Therefore the probability of getting one card from each suite is (¹⁰C₁)⁴/(⁴⁰C₄)

Example 08:

Find the probability that when a hand of 7 cards is dealt from a well-shuffled deck of 52 cards, it contains

Solution:

There are 52 cards in a pack.

Seven cards out of 52 can be drawn by ⁵²C₇ ways

Hence n(S) = ⁵²C₇

a) all 4 kings

Let A be the event of getting all 4 kings i.e. 4 kings and 3 non-king cards.

There are 4 kings and 48 non-king cards in a pack

∴ n(A) = ⁴C₄ x ⁴⁸C₃  

By the definition P(A) = n(A)/n(S) = (⁴C₄ x ⁴⁸C₃ )/(⁵²C₇)

Therefore the probability of getting all 4 kings is

(⁴C₄ x ⁴⁸C₃ )/(⁵²C₇)

b) exactly 3 kings

Let B be the event of getting exactly 3 kings i.e. 3 kings and 4 non-king cards.

There are 4 kings and 48 non-king cards in a pack

∴ n(A) = ⁴C₃ x ⁴⁸C₄  

By the definition P(A) = n(A)/n(S) = (⁴C₃ x ⁴⁸C₄ )/(⁵²C₇)

Therefore the probability of getting all 4 kings is

(⁴C₃ x ⁴⁸C₄ )/(⁵²C₇)

c) at least three kings

Let C be the event of getting at least three kings

There are two possibilities

Case – 1: Getting 3 kings and 4 non-king cards

Case – 2: Getting 4 kings and 3 non-king cards

There are 4 kings and 48 non-king cards in a pack

∴ n(C) = ⁴C₃ x  ⁴⁸C₄ + ⁴C₄ x  ⁴⁸C₃ 

By the definition P(C) = n(C)/n(S) = (⁴C₃ x  ⁴⁸C₄ + ⁴C₄ x  ⁴⁸C₃) / (⁵²C₇)

Therefore the probability of getting at least two face cards is

(⁴C₃ x  ⁴⁸C₄ + ⁴C₄ x  ⁴⁸C₃) / (⁵²C₇)

In the next article, we shall study some basic problems of probability based on the drawing of a single ball from a collection of identical coloured balls.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing 4 Playing Cards appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of three playing cards. For e.g. three cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting all red cards

Problems based on the Draw of 3 Playing Cards:

Example – 01:

Three cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

Three cards out of 52 can be drawn by ⁵²C₃ ways

Hence n(S) = ⁵²C₃ = 26 x 17 x 50

a) all face cards

Let A be the event of getting all face cards

There are 12 face cards in a pack

three face cards out of 12 can be drawn by ¹²C₃ ways

∴ n(A) = ¹²C₃ = 4 x 11 x 5

By the definition P(A) = n(A)/n(S) = (4 x 11 x 5)/( 26 x 17 x 50) = 11/1105

Therefore the probability of getting all face cards is 11/1105

b) no face card

Let B be the event of getting no face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

three non-face cards out of 40 can be drawn by ⁴⁰C₃ ways

∴ n(B) = ⁴⁰C₃ = 20 x 13 x 38

By the definition P(B) = n(B)/n(S) = (20 x 13 x 38)/( 26 x 17 x 50) = 38/85

Therefore the probability of getting no face card is 38/85.

c) atleast one face card

Let C be the event of getting at least one face card

Thus C is an event of getting no face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

three non-face cards out of 40 can be drawn by ⁴⁰C₃ ways

∴ n(C) = ⁴⁰C₃ = 20 x 13 x 38

By the definition P(C) = n(C)/n(S) = (20 x 13 x 38)/( 26 x 17 x 50) = 38/85

Now P(C) = 1 – P(C) = 1 – 38/85 = 47/85

Therefore the probability of getting at least one face card is 47/85.

d) at least two face cards

Let D be the event of getting at least two face cards

There are two possibilities

Case – 1: Getting two face cards and 1 non-face card

Case – 2: All three face cards

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

∴ n(D) = ¹²C₂ x  ⁴⁰C₁ + ¹²C₃  = 6 x 11 x 40 + 4 x 11 x 5 = 2860

By the definition P(D) = n(D)/n(S) = 3794/( 26 x 17 x 50) = 11/85

Therefore the probability of getting at least two face cards is 11/85.

e) at most two face cards

Let E be the event of getting at most two face cards

There are three possibilities

Case – 1: Getting no face card

Case – 2: Getting one face card and 2 non-face cards

Case – 3: Getting two face cards and 1 non-face card

There are 12 face cards in a pack

Thus there are 40 non-face cards in a pack

∴ n(E) = ⁴⁰C₃ +¹²C₁ x  ⁴⁰C₂ +  ¹²C₂ x  ⁴⁰C₁ = 20 x 13 x 38 + 12 x 20 x 39 + 6 x 11 x 40 = 21880

By the definition P(E) = n(E)/n(S) = 21880/( 26 x 17 x 50) = 1094/1105

Therefore the probability of getting at most two face cards is 1094/1105

f) all red cards

Let F be the event of getting all red cards

There are 26 red cards in a pack

three red cards out of 26 red cards can be drawn by ²⁶C₃ ways

∴ n(E) = ²⁶C₃ = 4 x 11 x 5 = 13 x 25 x 8

By the definition P(E) = n(E)/n(S) = (13 x 25 x 8)/( 26 x 17 x 50) = 2/17

Therefore the probability of getting all red cards is 2/17

f) all are not heart

Let F be the event of getting draw such that all are not heart

Thus F is the event that the draw consists of atmost two heart

There are three possibilities

Case – 1: Getting no heart

Case – 2: Getting one heart and 2 non hearts

Case – 3: Getting two hearts and 1 non heart

There are 13 heart cards in a pack

Thus there are 39 non-heart cards in a pack

∴ n(F) = ³⁹C₃ +¹³C₁ x  ³⁹C₂ +  ¹³C₂ x  ³⁹C₁ = 13 x 19 x 37 + 13 x 39 x 19 + 13 x 6 x 39 = 21814

By the definition P(F) = n(F)/n(S) = 21814/( 26 x 17 x 50) = 839/850

Therefore the probability of getting all not heart is 839/850

g) atleast one heart

Let G be the event of getting at least one heart

Thus G’ is an event of getting no heart

There are 13 heart cards in a pack

Thus there are 39 non-heart cards in a pack

three non-heart cards out of 39 non-heart cards can be drawn by ³⁹C₃ ways

∴ n(G’) = ³⁹C₃ = 13 x 19 x 37

By the definition P(G’) = n(G’)/n(S) = (13 x 19 x 37)/( 26 x 17 x 50) = 703/1700

Now P(G) = 1 – P(G’) = 1 – 703/1700 = 997/1700

Therefore the probability of getting at least one heart is 997/1700

h) a king,  a queen, and a jack

Let H be the event of getting a king,  a queen, and a jack

There are 4 kings, 4 queens and 4 jacks in a pack

Each specific selection can be done by ⁴C₁ ways

∴ n(H) = ⁴C₁ x ⁴C₁ x ⁴C₁ = 4 x 4 x 4 = 64

By the definition P(H) = n(H)/n(S) =64/( 26 x 17 x 50) = 16/5525

Therefore the probability of getting a king,  a queen and a jack is 16/5525

i) 2 aces and 1 king

Let J be the event of getting two aces and 1 king

There are 4 aces and 4 kings

∴ n(J) = ⁴C₂ x ⁴C₁  = 6 x 4  = 24

By the definition P(J) = n(J)/n(S) =24/( 26 x 17 x 50) = 6/5525

Therefore the probability of getting two aces and one king is 6/5525

In the next article, we shall study some basic problems of probability based on the drawing of four or more cards from a pack of playing cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing 3 Playing Cards appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving the draw of two playing cards. For e.g. Two cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting both red cards

Problems based on the Draw of 2 Playing Cards:

Example – 01:

Two cards are drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

two cards out of 52 can be drawn by ⁵²C₂ ways

Hence n(S) = ⁵²C₂ = 26 x 51

a) both club cards

Let A be the event of getting both club cards

There are 13 club cards in a pack

two club cards out of 13 club cards can be drawn by ¹³C₂ ways

∴ n(A) = ¹³C₂ =  13 x 6

By the definition P(A) = n(A)/n(S) = (13 x 6)/(26 x 51) = 1/17

Therefore the probability of getting both club cards is 1/17

b) both red cards

Let B be the event of getting both red cards

There are 26 red cards in a pack

two red cards out of 26 red cards can be drawn by ²⁶C₂ ways

∴ n(B) = ²⁶C₂ =  13 x 25

By the definition P(B) = n(B)/n(S) = (13 x 25)/(26 x 51) = 25/102

Therefore the probability of getting both red cards is 25/102

c) both black cards

Let C be the event of getting both black cards

There are 26 black cards in a pack

two black cards out of 26 black cards can be drawn by ²⁶C₂ ways

∴ n(C) = ²⁶C₂ =  13 x 25

By the definition P(C) = n(C)/n(S) = (13 x 25)/(26 x 51) = 25/102

Therefore the probability of getting both black cards is 25/102

d) both kings

Let D be the event of getting both kings

There are 4 kings in a pack

two kings out of four kings can be drawn by ⁴C₂ ways

∴ n(D) = ⁴C₂ =  2 x 3

By the definition P(D) = n(D)/n(S) = (2 x 3)/(26 x 51) = 1/221

Therefore the probability of getting both kings is 1/221

Note: The probability of getting two cards of a particular denomination is always 1/221

e) both red aces

Let E be the event of getting both red aces

There are 2 red aces in a pack

two red aces out of two red aces can be drawn by ²C₂ ways

∴ n(E) = ²C₂ =  1

By the definition P(E) = n(E)/n(S) = (1)/(26 x 51) = 1/1326

Therefore the probability of getting both red aces is 1/1326

f) both face cards

Let F be the event of getting both face cards

There are 12 face cards in a pack

two face cards out of 12 face cards can be drawn by ¹²C₂ ways

∴ n(F) = ¹²C₂ =  6 x 11

By the definition P(F) = n(F)/n(S) = (6 x 11)/(26 x 51) = 11/221

Therefore the probability of getting both red aces is 1/1326

g) cards of denomination between 4 and 10

Let G be the event of getting cards of denomination between 4 and 10

Denominations between 4 and 10 are 5, 6, 7, 8, 9 (total 5 denominations)

Each denomination has 4 cards

Thus there are 5 x 4 = 20 cards of denomination between 4 and 10

two such cards out of 20 can be drawn by ²⁰C₂ ways

∴ n(G) = ²⁰C₂ =  10 x 19

By the definition P(G) = n(G)/n(S) = (10 x 19)/(26 x 51) = 95/663

Therefore the probability of getting cards of denomination between 4 and 10 is 95/663

h) both red face cards

Let H be the event of getting both red face cards

There are 6 red face cards in a pack

two red face cards out of 6 red face cards can be drawn by ⁶C₂ ways

∴ n(H) = ⁶C₂ =  3 x 5

By the definition P(H) = n(H)/n(S) = (3 x 5)/(26 x 51) = 5/442

Therefore the probability of getting both red face cards is 5/442.

i) a queen and a king

Let J be the event of getting a queen and a king

There 4 kings and 4 queens in a pack

one king out of 4 can be selected by  ⁴C₁ ways and

one queen out of 4 can be selected by  ⁴C₁ ways

∴ n(J) = ⁴C₁ x ⁴C₁ =  4 x 4 = 16

By the definition P(J) = n(J)/n(S) = 16/(26 x 51) = 6/663

Therefore the probability of getting a queen and a king is 6/663

j) one spade card and another non-spade card.

Let K be the event of getting one spade card and another non-spade card.

There 13 spade cards and 39 non-spade cards in a pack

one spade card out of 13 spade cards can be selected by  ¹³C₁ ways and

one non-spade card out of 39 non-spade cards can be selected by  ³⁹C₁ ways

∴ n(K) = ¹³C₁ x ³⁹C₁ =  13 x 39

By the definition P(K) = n(K)/n(S) = (13 x 39)/(26 x 51) = 13/34

Therefore the probability of getting one spade card and another non-spade card is 13/34

l) both cards from the same suite

Let L be the event of getting both cards from the same suite

There 13 cards in each suite

two cards out of 13 cards of the same suite can be selected by  ¹³C₂ ways

There are four suites in a pack

Event M = both are spade cards or both are club cards or both are diamond cards or both are heart cards

∴ n(M) = ¹³C₂ + ¹³C₂ + ¹³C₂ + ¹³C₂ = 4 x ¹³C₂  = 4 x 13 x 6

By the definition P(M) = n(M)/n(S) = (4 x 13 x 6)/(26 x 51) = 4/17

Therefore the probability of getting both cards of the same suite is 4/17

m) both are of the same denomination

Let N be the event of getting both cards of the same denomination

There 4 cards of the same denomination

two cards out of 4 cards of the same denomination can be selected by  ⁴C₂ ways

There are 13 sets of the same denomination

∴ n(N) =  13 x ⁴C₂  = 13 x 2 x 3

By the definition P(N) = n(N)/n(S) = (13 x 2 x 3)/(26 x 51) = 1/17

Therefore the probability of getting both cards of the same denomination is 1/17

o) One is spade and other is ace

Let Q be the event of getting one spade and another ace

There are two possibilities

Case – 1: When the first card is spade with spade ace included and another is ace from remaining three aces

Case – 2: When the first card is spade with ace excluded and another is ace from four aces

∴ n(Q) =  ¹³C₁ x ³C₁  +  ¹²C₁ x ⁴C₁  = 13 x 3 + 12 x 4 = 39 + 48 = 87

By the definition P(N) = n(N)/n(S) = 87/(26 x 51) = 29/442

Therefore the probability of getting one spade and other ace is 29/442

In the next article, we shall study some basic problems of probability based on the drawing of three cards from a pack of playing cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing 2 Playing Cards appeared first on The Fact Factor.

In the last few articles, we have seen to solve problems based on tossing of coins, throwing dice, and selecting numbered cards. In this article, we shall study the problems to find the probability involving playing cards.

Introduction to Playing Cards:

Before studying, the problems on playing cards, you should be thorough with the following facts:

• There are 52 playing cards in a pack of playing cards.
• There are four suites in a pack viz: Spade (♠), Club (♣), Diamond (♦), Heart (♥)
• In each suite, there are 13 cards of different Denominations. viz. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen and King
• Thus there are 4 cards of each denomination in a pack, like 4 kings, 4 queens, 4 aces, 4 tens, 4 fives, etc.
• Spade and Club are black cards while Diamond and Heart are red cards.
• There are 26 black cards and 26 red cards in a pack.
• Each card is unique in a pack.
• King, Queen, and Jack cards are called picture cards or face cards.
• Thus there are total 12 face cards in a pack. 6 black face cards, 6 red face cards in a pack of playing cards
• There are 3 face cards in each suite.
• The Ace, King, Queen, and Jack of each suit are called honour cards
• The rest of the cards (2, 3, 4, 5, 6, 7, 8, 9, 10 ) are called spot cards.
• Spades and Hearts are called the major suits and Diamonds and Clubs are called the minor suits

Drawing a Single Playing Card From a Pack:

Example – 01:

A card is drawn from a well-shuffled pack of 52 playing cards. Find the probability of getting

Solution:

There are 52 cards in a pack.

one card out of 52 can be drawn by ⁵²C₁ ways

Hence n(S) = ⁵²C₁ = 52

a) a spade card

Let A be the event of getting a spade card

There are 13 spade cards in a pack

one spade card out of 13 can be drawn by ¹³C₁ ways

∴ n(A) = ¹³C₁ =  13

By the definition P(A) = n(A)/n(S) = 13/52 = 1/4

Therefore the probability of getting a spade card is 1/4

b) a red card

Let B be the event of getting a red card

There are 26 red cards in a pack

one red card out of 26 can be drawn by ²⁶C₁ ways

∴ n(A) = ²⁶C₁ =  26

By the definition P(B) = n(B)/n(S) = 26/52 = 1/2

Therefore the probability of getting a red card is 1/2

c) a black card

Let C be the event of getting a black card

There are 26 black cards in a pack

one black card out of 26 can be drawn by ²⁶C₁ ways

∴ n(C) = ²⁶C₁ =  26

By the definition P(C) = n(C)/n(S) = 26/52 = 1/2

Therefore the probability of getting a black card is 1/2

d) a king

Let D be the event of getting a king

There are 4 kings in a pack

one king out of 4 can be drawn by ⁴C₁ ways

∴ n(D) = ⁴C₁ =  4

By the definition P(D) = n(D)/n(S) = 4/52 = 1/13

Therefore the probability of getting a king is 1/13

Note: Probability of getting a card of a particular denomination is always 1/13

e) a red ace

Let E be the event of getting a red ace

There are 2 red aces in a pack

one red ace out of 2 can be drawn by ²C₁ ways

∴ n(E) = ²C₁ =  2

By the definition P(E) = n(E)/n(S) = 2/52 = 1/26

Therefore the probability of getting a red ace is 1/26

f) a face card

Let F be the event of getting a face card

There are 12 face cards in a pack

one face card out of 12 can be drawn by ¹²C₁ ways

∴ n(F) = ¹²C₁ =  12

By the definition P(F) = n(F)/n(S) = 22/52 = 3/13

Therefore the probability of getting a face card is 3/13

g) a card of denomination between 4 and 10

Let G be the event of getting a card of denomination between 4 and 10

Denominations between 4 and 10 are 5, 6, 7, 8, 9 (total 5 denominations)

Each denomination has 4 cards

Thus there are 5 x 4 = 20 cards of denomination between 4 and 10

one such card out of 20 can be drawn by ²⁰C₁ ways

∴ n(G) = ²⁰C₁ =  20

By the definition P(G) = n(G)/n(S) = 20/52 = 5/13

Therefore the probability of getting a card of denomination between 4 and 10 is 5/13

h) a red face card

Let H be the event of getting a red face card

There are 6 red face cards in a pack

one face card out of 6 can be drawn by ⁶C₁ ways

∴ n(G) = ⁶C₁ =  6

By the definition P(H) = n(H)/n(S) = 6/52 = 3/26

Therefore the probability of getting a red face card is 3/26.

i) a queen of hearts

Let J be the event of getting a queen of hearts

There is only one queen of heart in a pack

one queen of hearts out of 1 can be drawn by 1way

∴ n(J) = 1

By the definition P(J) = n(J)/n(S) = 1/52

Therefore the probability of getting a queen of hearts is 1/52

j) a queen or a king

Let K be the event of getting a queen or a king

There 4 kings and 4 queens in a pack

Thus there are 4 + 4 = 8 favourable points.

one required card out of 8 favourable points can be drawn by ⁸C₁ ways

∴ n(K) = ⁸C₁ =  8

By the definition P(K) = n(K)/n(S) = 8/52 = 2/13

Therefore the probability of getting a queen or a king is 2/13

k) a red card and a king

Let L be the event of getting a red card or a king

There 2 red cards which are king

Thus there are 2 favourable points.

one required card out of 2 favourable points can be drawn by ²C₁ ways

∴ n(L) = ²C₁ =  2

By the definition P(L) = n(L)/n(S) = 2/52 = 1/26

Therefore the probability of getting a red card and king is 1/26

l) a red card or a king  /a red king

Let M be the event of getting a red card or a king

There 26 red cards (including 2 red kings) and 2 black kings in a pack

Thus there are 26 + 2 = 28 favourable points.

one required card out of 28 favourable points can be drawn by ²⁸C₁ ways

∴ n(M) = ²⁸C₁ =  28

By the definition P(M) = n(M)/n(S) = 28/52 = 7/13

Therefore the probability of getting a red card or a king (a red king) is 7/13

m) Neither the heart nor the king

Let N be the event of getting neither the heart nor the king

There 36 non-heart cards (excluding 3 kings) in a pack

one required card out of 36  favourable points can be drawn by ³⁶C₁ ways

∴ n(N) = ³⁶C₁ =  36

By the definition P(N) = n(N)/n(S) = 36/52 = 9/13

Therefore the probability of getting neither the heart nor the king is 9/13

n) Neither an ace nor the king

Let Q be the event of getting neither an ace nor a king

There are 4 aces and 4 kings in a pack

There 44 non-ace and non-king cards in a pack

one required card out of 44  favourable points can be drawn by ⁴⁴C₁ ways

∴ n(Q) = ⁴⁴C₁ =  44

By the definition P(Q) = n(Q)/n(S) = 44/52 = 11/13

Therefore the probability of getting neither ace nor the king is 11/13

o) no diamond

Let R be the event of getting no diamond

There 39 non-diamond cards in a pack

one required card out of 39  favourable points can be drawn by ³⁹C₁ ways

∴ n(R) = ³⁹C₁ =  39

By the definition P(R) = n(R)/n(S) = 39/52 = 3/4

Therefore the probability of getting no diamond is 3/4

p) no ace

Let T be the event of getting no ace

There are 4 aces in a pack

There 48 non-ace cards in a pack

one required card out of 48  favourable points can be drawn by ⁴⁸C₁ ways

∴ n(T) = ⁴⁸C₁ =  48

By the definition P(T) = n(T)/n(S) = 48/52 = 12/13

Therefore the probability of getting no ace is 12/13.

q) not a black card

Let V be the event of getting no black card

There 26 non-black (red) cards in a pack

one required card out of 26  favourable points can be drawn by ²⁶C₁ ways

∴ n(V) = ²⁶C₁ =  26

By the definition P(V) = n(V)/n(S) = 26/52 = 1/2

Therefore the probability of getting no black card is 1/2.

In the next article, we shall study some basic problems of probability based on the drawing of two cards from a pack of playing cards.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Drawing a Playing Card appeared first on The Fact Factor.

When a human blood sample is prevented from clotting and spun in a test tube (centrifuged), in a machine called a centrifuge, the blood separates into a   straw coloured liquid called plasma and a dark brown mass of blood cells. The lower layer consists of white blood cells, blood platelets, and red blood cells. Collectively, these are the formed elements, which make up about 45% of the total volume of whole blood; the percentage of blood attributed to red blood cells is called the hematocrit. The hematocrit is defined as the percentage of blood volume that is occupied by erythrocytes. The normal hematocrit is approximately 45 percent in men and 42 percent in women.

The upper layer is plasma, which contains a variety of inorganic and organic molecules dissolved or suspended in water. Plasma accounts for about 55% of the total volume of whole blood.

Composition of Plasma:

Plasma is the straw-coloured non-living, liquid part of blood. It makes up about 55 – 60% of blood volume and 5.5 % of body weight. Blood corpuscles and platelets are suspended in it. Blood without clotting factor is called serum. The characteristic straw color of plasma is due largely to a waste product of hemoglobin breakdown called bilirubin.

It is the liquid portion of blood, and about 92% of plasma is water. The remaining 8% of plasma is composed of various salts (ions) and organic molecules. The salts, which are dissolved in plasma, help maintain the pH of the blood. Small organic molecules such as glucose, amino acids, and urea are also dissolved in plasma. The large organic molecules in plasma include hormones and the plasma proteins.

Water:

Plasma is composed of about 90 to 92% water. Acts as a solvent and suspending medium for blood components.  

Plasma Proteins:

Plasma proteins or serum proteins constitute 6 to 8% of plasma. Important plasma-proteins are. The plasma proteins constitute, by weight, most of the plasma solutes. They can be classified, according to certain physical and chemical reactions, into three broad groups: the albumins and globulins, and fibrinogen, which function in blood clotting. Most plasma proteins are made in the liver. An exception is the antibodies produced by B lymphocytes, which function in immunity. Certain hormones are plasma proteins made by various glands.  It must be emphasized that the plasma proteins normally are not taken up by cells; cells use plasma amino acids, not plasma proteins, to make their own proteins. Plasma proteins must be viewed quite differently from most of the other organic constituents of plasma, which use the plasma as a medium for transport to and from cells. In contrast, most plasma proteins perform their functions in the plasma itself or in the interstitial fluid.

• Fibrinogen and prothrombin: It constitutes 4% of the plasma proteins and required for blood clotting.
• Serum albumin: The albumins are the most abundant of the three plasma protein groups and are synthesized by the liver. It makes up 58% of the plasma proteins. They are partly responsible for blood viscosity, the regulation of water movement between tissues and blood and osmotic pressure; acts as a buffer; transports fatty acids,
  free bilirubin, and thyroid hormones.
• Globulins or Gamma globulins or Immunoglobins (Ig): It accounts for 38% of the plasma proteins. They act as antibodies and are associated with the defence mechanism of the body. They Transport lipids, carbohydrates, hormones, and ions like iron and copper; antibodies complement are involved in immunity

Inorganic Salts and Ions (Minerals):

They form 1-2 % of the plasma and includes electrolytes like Sodium, potassium, calcium, magnesium, chloride, iron, phosphate, hydrogen, hydroxide, bicarbonate. They are involved in osmosis, membrane potentials, and acid-base balance.

Dissolved Nutrients:

Glucose, lipids, vitamins, fatty acids, amino acids, and cholesterol. They act as sources of energy and basic “building blocks” of more complex molecules. Vitamins Promote enzyme activity.

Hormones and Enzymes:

Blood acts as the transport system for the transportation of regulatory substances called hormones secreted by different glands.  Thus plasma contains hormones and enzymes. Enzymes catalyze chemical reactions; hormones stimulate or inhibit many body functions.

Dissolved Gases:

Oxygen, It is necessary for aerobic respiration; terminal electron acceptor in an electron-transport chain, Carbon dioxide, a Waste product of aerobic respiration; as bicarbonate, helps buffer blood and nitrogen.

Excretory Substances:

Ammonia, urea, uric acid, creatine, and creatinine. Urea, uric acid, creatinine , and ammonia salts  are the breakdown products of protein metabolism; excreted by the kidneys. Bilirubin is the breakdown product of red blood cells; excreted as part of the bile from the liver into the intestine. Lactic acid is the end product of anaerobic respiration; converted to glucose by the liver.

Functions of Plasma:

Transport Nutrients:

Delivering nutrients to the body is a critical function of the circulatory system. Plasma of the blood is the carrier of all nutrients. After food is digested and assimilated, its component nutrients like carbohydrates, proteins, minerals, fats, and vitamins are absorbed into the bloodstream. Each of these nutrients is vital for healthy body function.

Transport of Waste Products:

The plasma collects metabolic waste products like urea, creatinine, and other chemical wastes and toxins and transports them to the liver, kidneys, skin, and lungs (excretory organs) for elimination from the body.

Transport of Hormones:

Hormones are chemical messengers produced by endocrine glands that affect distant organs. Hormones are released into the bloodstream through which they travel to target sites. The plasma collects the hormones from the endocrine glands and serves as the transportation connection between the glands and the organs or tissues.

Transport of Other Products:

Albumin transports the molecule bilirubin, a breakdown product of hemoglobin. Lipoproteins, whose protein portion is a globulin, transport cholesterol.

Body Temperature Regulation:

Plasma Picks up excess body heat from the deep-seated heat-producing organs and brings it to the skin to be excreted. If body temperature drops, surface blood vessels constrict(decrease in size) to conserve body heat. Thus it  helps in regulating the body temperature

Disease Protection and Healing:

There are three types of globulins, designated alpha, beta, and gamma globulins. The alpha and beta globulins, produced by the liver, bind to metal ions, to fat-soluble vitamins, and to lipids, forming the lipoproteins. Antibodies, which help fight infections by combining with antigens, are gamma globulins. The immunoglobins of plasma act as antibodies and attack the foreign intruder in the body. They neutralize these harmful foreign agents. Thus plasma is responsible for the immunity of the body.

Fibrinogen present in the plasma is responsible for clotting of blood which is important for stopping the blood flow from the wounds.

Maintain Haemostasis and Osmoregulation:

Plasma supplies water to different tissues at the same time and removes excess of water produced during metabolic activities. Thus it maintains water balance in the body. Osmotic pressure is a force caused by a difference in solute concentration on either side of a membrane. The plasma proteins, particularly the albumins, contribute to the osmotic pressure, which pulls water into the blood and helps keep it there.

Acid-Base Buffer:

Plasma proteins act as acid-base buffers and maintain blood pH within a range. Plasmaproteins are able to take up and release hydrogen ions; therefore, the plasma proteins help buffer the blood and keep its pH around 7.40.

Disorders Related with Blood Plasma:

Oedema or Edema:

Oedema is swelling that occurs when too much fluid becomes trapped in the tissues of the body, particularly the skin. It most often occurs in the skin, especially on the hands, arms, ankles, legs, and feet. However, it can also affect the muscles, bowels, lungs, eyes, and brain. It usually starts slowly, but the onset can be sudden.

In case of a person suffering from protein deficiency, a fall in plasma protein leads to escape of excess volume of water from the blood to tissues. Due to excess of fluid of fluid in tissues causes swelling of feet. The state is called oedema.  The condition mainly occurs in older adults and women who are pregnant. Symptoms include skin that retains a dimple after being pressed for a few seconds,  puffiness of the ankles, face, or eyes, higher pulse rate and high blood pressure.

Diuretics are a type of medication. They help get rid of excess fluid by increasing the rate of urine production by the kidneys.

Cholesterol:

Cholesterol is present in plasma. Cholesterol has a tendency to deposit on the walls of blood vessels leading to a condition called atherosclerosis. The liver is responsible for producing and clearing cholesterol in the body.

Dietary cholesterol increases plasma total cholesterol concentrations in humans. There is a relationship between increased plasma cholesterol concentrations and cardiovascular disease risk. Dietary guidelines have consistently recommended to such person limiting food sources of cholesterol. Potential sources of dietary cholesterol are limited to animal foods; eggs, dairy products, and meat.

LDL (low-density lipoprotein) cholesterol is also called “bad” cholesterol. LDL can build up on the walls of arteries and increase the chances of getting heart disease. HDL (high-density lipoprotein) cholesterol is also called “good” cholesterol. HDL protects against heart disease by taking the bad cholesterol out of the blood and keeping it from building up in arteries. Along with cholesterol, triglycerides form plasma lipids. Excess triglycerides in plasma have been linked to the occurrence of coronary artery disease in some people.

Everyone over the age of 20 should get their cholesterol levels checked at least once every 5 years by a test called “Lipid profile”. Everyone over the age of 40 should get their cholesterol levels checked at least once a year.

Lifestyle changes such as exercising and eating a healthy diet are the first line of defence against high cholesterol. The choice of medication for high cholesterol depends on individual risk factors, age,  current health and possible side effects. Common choices include Statins, Bile-acid-binding resins, Cholesterol absorption inhibitors, Injectable medications. The choice of medication for high triglycerides is Fibrates, Niacin, Omega-3 fatty acid supplements.

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Science > Biology > Human Anatomy and Physiology > Cardiovascular System > Composition of Blood: Blood Plasma

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In the Cardiovascular system, the ‘‘heart’’ (cardi) pumps the blood in a ‘‘little circle’’ (circul), which travels through ‘‘little vessels’’ (vascul). In human, the transportation is done through blood circulatory system and lymphatic system. Thus two fluids move through the circulatory system: blood and lymph. The blood, heart, and blood vessels form the Cardiovascular System. The lymph, lymph nodes and lymph vessels form the Lymphatic System. Human blood circulatory system has three main components. A fluid (blood), tubing (arteries, veins and capillaries) and a pump (the heart).

Study of blood is called haematology. It is a fluid connective tissue. It is bright red, slightly alkaline, salty viscous fluid heavier than water.

The Constituent of the Blood:

In this article, we shall only take an overview of the composition. In the next article, each constituent is discussed in detail.

When a human blood sample is prevented from clotting and spun in a test tube (centrifuged), in a machine called a centrifuge, the blood separates into a   straw coloured liquid called plasma and a dark brown mass of blood cells. The lower layer consists of white blood cells, blood platelets, and red blood cells. Collectively, these are the formed elements, which make up about 45% of the total volume of whole blood; the percentage of blood attributed to red blood cells is called the hematocrit. The hematocrit is defined as the percentage of blood volume that is occupied by erythrocytes. The normal hematocrit is approximately 45 percent in men and 42 percent in women.

The upper layer is plasma, which contains a variety of inorganic and organic molecules dissolved or suspended in water. Plasma accounts for about 55% of the total volume of whole blood.

Characteristics of Human blood:

Study of blood is called haematology. Blood is a fluid connective tissue. It is bright red, slightly alkaline (pH 7.3 to 7.5), salty viscous fluid heavier than water. the pH of blood is more in arteries than that in veins. The viscosity of blood is 5 to 6 times that of water. An adult has a blood volume of approximately 5.5 litres. It forms 6 to 10 % of the body weight. Blood is the only tissue that exists in both the liquid and solid state simultaneously.

Volume of Blood and Different Constituents:

The volume of blood in an average-sized person is approximately 5.5 L. Now hematocrit is 45 percent of the total volume,

Then, Erythrocyte volume = 0.45 x 5.5 L = 2.5 L

Since the volume occupied by the leukocytes and platelets is normally considered negligible, the plasma volume equals the difference between blood volume and erythrocyte volume; therefore, in our average person

Plasma volume = 5.5 L – 2.5 L = 3.0 L

Plasma:

The plasma consists of 90 to 92% water and 8% to 10% proteins, salts, hormones, enzymes, waste products and other various chemicals. Most of the solute part about 7% is proteins. These include antibodies that help to protect the body from diseases, fibrinogen that helps the blood to clot. The waste product includes urea and carbon dioxide. Hormones are the chemical messenger, which help to coordinate different body functions.

Plasma obtained from blood donation may be converted to a powdered form for storage. During the transfusion, it is dissolved in sterile distilled water and can be administered at once. This method saved the lives of many during World war.

Red Blood Cells (RBCs): 

They are also called erythrocyte. They are produced inside the bone marrow. They have a lifespan of about 100 to 120 days after which they are destroyed by the liver. They are the most common type of blood cell (5.1 to 5.8 million per cubic mm). They are non-nucleated, small in size, round and biconvex in shape. They are able to fold and bend as they are forced through the smallest blood vessels.

The strong red colour of blood is due to the large number of RBCs. Red blood cells contain haemoglobin which gives them their red colour and enables them to carry oxygen from lungs to different parts of the body. They also carry carbon dioxide from different parts of the body to the lungs. Their main function is to carry oxygen from the lungs to the tissues.

White blood cells (WBCs):

They are also called leucocytes. Many white cells are made in the bone marrow. Their lifespan is 3 to 4 days. They are colourless and they have a nucleus. WBCs are larger than red cells but they are lesser in number. (About 5000-7000 per cubic millimetre of blood). They are further classified as lymphocytes and phagocytes. The nucleus of each type has a characteristic shape. When they travel in the blood they are more or less spherical, but they flatten and continuously change their shape along the inside walls of the blood vessels.

WBC’s rid the body of pathogens in the process called phagocytosis. In this process, the WBC surrounds, engulfs and “eats” the invading pathogen.

Platelets : 

Platelets are also called thrombocytes. They are made in the bone marrow and have a lifespan of 8-14 days. They are much smaller than red cells. One cubic millimetre of blood contains about a quarter of million platelets. Their function is to help the blood clot. Clotting prevents loss of blood from wounds.

Observing Blood Under Microscope:

• Clean the skin of your finger with a swab of cotton dipped in ethanol. With a sterile needle prick your finger so that a drop of blood comes out.
• Place the drop of blood at one end of a microscope slide. With another slide spread the blood over the surface to form a smear. Let it dry and then examine it under a microscope. We can observe red blood corpuscles.
• Now cover the smear with Leishman’s stain and leave it for five minutes. Then wash the stain off with tap water gently. Let the slide dry and then examine it under the microscope again. Now, we can observe white blood corpuscles.

Functions of Human blood:

Transportation:

The blood moves from the heart to all the various organs, where exchange with tissues takes place across thin capillary walls. Blood collects oxygen from the lungs and nutrients from the digestive tract and transports these to the tissues. It delivers enzymes and chemical messengers to cells and tissues. Various organs and tissues secrete hormones into the blood, and blood transports these to other organs and tissues, where they serve as signals that influence cellular metabolism. It delivers water, vitamins and minerals to cells. It carries carbon dioxide from cells, tissues and carries it to lungs for disposal. It carries waste materials like urea and other chemical wastes and carries them to the liver and kidneys for disposal. It carries antibodies from place to place in the body. It carries vitamins and enzymes.

Protection:

The blood defends the body against invasion by pathogens (microscopic infectious agents, such as bacteria and viruses) in several ways. Certain blood cells are capable of engulfing and destroying pathogens, and others produce and secrete antibodies into the blood. White blood corpuscles  (WBC) fight against disease-causing germs that harm the body.

It prevents the loss of blood after an injury by clotting. Blood helps in the repair process after a cut or other injury. Without blood clotting, we could
bleed to death even from a small cut.

Regulation:

Blood Picks up excess body heat and brings it to the skin to be excreted. The sweat is formed on the skin. Which is evaporated and heat required for it is taken from the body. Hence the body cools down. It controls the amount of water in the body. The salts and plasma proteins in blood act to keep the
liquid content of blood high. In this way, blood plays a role in helping to maintain its own water-salt balance. It also helps to regulate the amount of chemical substance in the tissues of the body. Due to the presence of buffers in the blood, it also helps to regulate body pH and keep it relatively constant.

Blood Donation:

Anybody who is healthy weighs over 50 kg and is between the age of 18 years and 65 years can donate blood. An adult has a blood volume of approximately 5 litres. A donor may give up to half a litre of blood at one time. This is quickly replaced by the body.

Many donors give blood regularly. It is immediately mixed with a chemical which prevents it from clotting and also provides food for the living cells. The blood is then stored in a refrigerator until it is required. Similarly, sodium citrate is added to it to avoid coagulation. The place where the blood is stored is called a blood bank.

Blood groups:

Human blood group is determined by the antigens present on the surface of RBC’s.  Blood groups are inherited and do not change throughout life. Human blood is classified into 4 main groups: A, B, AB and O. Each can be either Rhesus + ve or Rhesus – ve, giving  8 groups in all. Blood grouping is the identification of the antigens in a blood sample. This system is called ABO system.

An individual’s RBC’s may carry an A antigen, a B antigen, both A and B antigens, or no antigen at all. These antigen patterns are called blood types A, B, AB and O, respectively. Type AB is known as a universal recipient, meaning that they can receive any type of blood, while O is the universal donor, meaning they can donate blood to anyone

Blood transfusion:   

Blood transfusion is the transfer of blood that is taken from one person, into the bloodstream of another person. The person who gives blood is called the donor. The person who receives blood is called the recipient. When there is a loss of blood suddenly due to an accident, or because of the bursting of a blood vessel, there is a danger that not enough blood will be left to maintain the circulation. In such a case, the patient may lose consciousness due to low blood pressure and hence less supply of oxygen to tissues. Losing blood is called haemorrhage. To restore the blood volume and to provide more red cells, a blood transfusion is carried out.

Before doing blood transfusion the compatibility between the groups of the donor and the recipient should be checked. If the blood of the donor is not compatible with the blood of the patient, the red cells in the patient’s blood will stick together. This may lead to death.

Clotting of Blood:

Human blood contains heparin and antithrombins as anticoagulants, it prevents the blood to clot inside the blood vessels. As soon as blood vessel ruptures, bleeding starts. The conversion of liquid blood into semisolid jelly is called blood coagulation or blood clotting. Platelets adhere to the site of the wound and release clotting factors known as prothrombin. Prothrombin is inactive. At the site of rupture the platelets and injured tissues release thromboplastin which initiates the formation of enzyme prothrombinase.  In the presence of Ca, ions prothrombinase converts inactive prothrombin to active thrombin. Thrombin converts soluble fibrinogen into fibrin. The fibrin forms a net to enmesh platelets blood cells and plasma to form a clot.

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Science > Biology > Human Anatomy and Physiology > Cardiovascular System > The blood, an Overview

The post The Blood appeared first on The Fact Factor.

In living things, many substances such as food, gases, minerals salts, hormones and waste products have to be transported from one part of the body to another. Plants and animals have a system of transporting substances throughout their body. In animals (vertebrates), blood is the medium of transport. In small animals (lower animals) such as protozoa and Flatworms, Diffusion of gases takes place through their body surface. Due to their smaller size, they lack a special system for the transport of other materials such as oxygen for existence. Higher animals have higher metabolic activity, hence higher animals require an efficient and speedy supply of nutrients and oxygen for their tissues at the same time they require efficient and speedy disposal of respiratory products, nitrogen products. Hence such animals have developed a special fluid called blood in them and conducting system called circulatory system consisting of heart, vessels etc. This system is also called a vascular system. The circulatory system has two functional components viz. a) blood vascular system and b) lymph vascular system.

Functions of the Circulatory System

Transport of oxygen:

The main function of the circulatory system is supplying oxygen to the different parts of the body. Every cell requires oxygen for their survival. Particularly the brain cells are the most sensitive and begin to die in as little as 3 minutes if deprived of oxygen. During inhalation, air enters the lungs and oxygen is absorbed through the air sacs (alveoli) into the bloodstream. Oxygen combines with the haemoglobin of RBCs to form oxyhaemoglobin. This oxygen-rich blood is pumped through the heart into the arterial circulation. In the capillaries, unstable oxyhaemoglobin breaks down into haemoglobin and oxygen. The separated oxygen diffuses out of the blood into the cells of the body’s organs and tissues where it is utilized for metabolic activities.

Transportation of Carbon dioxide:

Carbon dioxide is a waste product produced by cells during its metabolic activities. A small amount of carbon dioxide combines with haemoglobin to for carbamino-haemoglobin. Some carbon dioxide dissolves in blood plasma.  The absorbed carbon dioxide in the blood is transported to the lungs through the venous circulation. When this oxygen-poor blood reaches the lungs, carbon dioxide diffuses through the air sacs (alveoli) and is then exhaled.

Transport of Nutrients:

Delivering nutrients to the body is another critical function of the cardiovascular system. After food is digested and assimilated, its component nutrients like carbohydrates, proteins, minerals, fats, and vitamins are absorbed into the bloodstream. Each of these nutrients is vital for healthy body function. Carbohydrates are a direct source of energy for the body while proteins (amino acids) are building blocks of new cells. Thus circulatory system supply nutrients wherever it is required. Like oxygen, nutrients diffuse from the bloodstream into body cells via the capillaries.

Transport of Waste Products:

The circulatory system collects metabolic waste products like urea and other chemical wastes and toxins and transports them to the liver, kidneys, skin, and lungs (excretory organs) for elimination from the body.

Transport of Hormones:

Hormones are chemical messengers produced by endocrine glands that affect distant organs. Hormones are released into the bloodstream through which they travel to target sites. Thus the cardiovascular system serves as the transportation connection between the endocrine glands and the organs or tissues. For example, Pituitary gland situated near the brain produces hormones which control other endocrine glands such as the thyroid, ovaries and testes and growth. Similarly, pancreas situated near liver produce hormone insulin which is required for maintaining blood sugar level throughout the body. The circulatory system delivers these hormones to the site of use of that hormone.

Body Temperature Regulation:

Blood Picks up excess body heat and brings it to the skin to be excreted. The sweat is formed on the skin. Which is evaporated and heat required for it is taken from the body. Hence the body cools down. Optimal function of the human body occurs within a relatively narrow temperature range, which is tightly regulated. If body temperature begins to rise, blood vessels close to the body surface dilate (increasing in size). This allows the body to get rid of an excess of heat through the skin. If body temperature drops, surface blood vessels constrict (decrease in size) to conserve body heat. Thus the cardiovascular system works in concert with the body’s sweating mechanism  to regulate the body temperature

Disease Protection and Healing:

The circulatory system serves as the path for disease-fighting cells and proteins, and messengers of the immune system. WBC’s rid the body of pathogens (invading germs) in the process called phagocytosis. In this process, the WBCs surround, engulf and “eats” the invading pathogen. Thus they fight against the pathogens. On entry of germs, antibodies are produced in the blood and a chemical alarm signal is created that travel through the bloodstream, which subsequently transports antibodies to the site of the infection. The circulatory system also carries chemical messengers that attract cells to heal tissues that have been damaged due to injury or disease. It helps in clotting to heal the wound.

Maintain Haemostasis and Osmoregulation:

Homeostasis is a property of cells, tissues, and organisms that allows the maintenance and regulation of the stability and constancy needed to function properly. Homeostasis is a healthy state that is maintained by the constant adjustment of biochemical and physiological pathways. Humans’ internal body temperature, the blood sugar level, blood pressure are all maintained through homeostasis. The circulatory system plays an important role in it. The maintenance of constant osmotic pressure in the fluids of an organism by the control of water and salt concentrations is called osmoregulation. The osmotic pressure and concentrations are maintained by the circulatory system.

Types of Circulatory Systems

Higher organisms have higher metabolic activity, hence higher animals require an efficient and speedy supply of nutrients and oxygen for their tissues at the same time they require efficient and speedy disposal of respiratory products, nitrogen products. Hence such animals have developed a special fluid called blood in them and conducting system called circulatory system consisting of heart, vessels etc. This system is also called a vascular system. The circulatory system has two functional components viz. a) blood vascular system and b) lymph vascular system.

The blood vascular system is of two types a) Open circulation and  b) Closed circulation

Open Circulatory System:

Open circulatory systems are the more basic type of circulatory system. In this circulatory system, the blood is not contained within an enclosed circuit of vessels. Blood (haemolymph) flows from the heart through open-ended vessels and, when it reaches the end of the vessels, The haemolymph enters into an open cavity called a haemocoel. The haemolymph mixes with interstitial fluid and moves around the haemocoel, thus bathing the internal organs and tissues. Haemolymph flows directly over the tissues delivering nutrients and in some cases, gases such as oxygen. Then the haemolymph freely flows back into vessels that direct the blood back to the heart.

The heart is simply an aorta or other blood vessels, and the haemolymph is pulsed throughout the body by muscle contractions. There are no arteries or major veins to pump the haemolymph, so blood pressure is very low and the volume of haemolymph is relatively high.

Examples: arthropods which include insects, spiders, prawns and most molluscs.

Characteristics of Open Circulatory System:

• Blood flows through large open spaces called lacunae and sinuses.
• No capillary system so tissues are in direct contact with haemolymph.
• Exchange of nutrients and gases takes place directly between haemolymph and tissue.
• The volume of haemolymph flowing through tissues cannot be controlled as hemolymph is flowing through open spaces.
• haemolymph flow is very slow. The blood pressure is very low
• Found in higher invertebrates like most arthropods such as prawns, cockroach etc. and in some molluscs.

Closed Circulatory System:

William Harvey (1578-1657)  discovered and published the first accurate description of the human circulatory system, based on his many years of experiments and observations as a scientist and physician. The humans have closed circulatory system. Unlike an open circulatory system, a closed circulatory system is more structured and controlled. The blood of a closed system always flows inside vessels. These vessels make up the vessel system of the body and can be found throughout the entire body.

Characteristics of Closed Circulatory System:

• Blood flows through a closed system i.e. heart and blood vessels.
• Capillary system present so blood is not in direct contact with tissues.
• Nutrients and gases pass through the walls of capillaries to tissue fluid which is then taken up by the tissues.
• Blood flow is controlled by contraction and relaxation of muscles of blood vessels.
• Blood flow is rapid here.
• Found in some molluscs, annelids and all vertebrates.

Importance of Human Circulatory System:

Most of the cells in the human body are not in direct contact with the external environment, so rely on the circulatory system to act as a transport service for them. Two fluids move through the circulatory system: blood and lymph. The blood, heart, and blood vessels form the Cardiovascular System. The lymph, lymph nodes and lymph vessels form the Lymphatic System. The Cardiovascular System and the Lymphatic System collectively make up the Circulatory System. The system in which blood is circulated throughout the body is called the circulatory system.

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