Problems Based on Selection of Balls

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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:

9 R

7W

4 B

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:

9 R

7 W

4 B

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.

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