In the last few articles, we have studied the basic concepts of probability. In this article, we are going to study problems based on the tossing of coins.

Algorithm:

1. Study experiment and write the sample space
2. Find favourable point and write event space
3. Use the definition of probability and find it.

Tossing of a Single Coin:

Example 01:

A fair coin is tossed once. Find the probability of getting

A fair coin is tossed.

The sample space is  S = {H, T}  and n(S) = 2

a) Getting a head

Let A be an event of getting head

A = {H} and n(A) = 1

By definition of probability

P(A) = n(A)/n(S) = 1/2

Ans: Thus the probability of getting head is 1/2

b) Getting a tail

Let B be an event of getting tail

B = {T} and n(B) = 1

By definition of probability

P(B) = n(B)/n(S) = 1/2

Ans: Thus the probability of getting tail is 1/2

c) Getting not the head

Let C be an event of getting not head

C = {T} and n(C) = 1

By definition of probability

P(C) = n(C)/n(S) = 1/2

Ans: Thus the probability of getting not head is 1/2

d) Getting not the tail

Let D be an event of getting not tail

D = {H} and n(D) = 1

By definition of probability

P(D) = n(D)/n(S) = 1/2

Ans: Thus the probability of getting not tail is 1/2

e) Getting the head or the tail

Let E be an event of getting the head or the tail

E = {H, T} and n(E) = 2

By definition of probability

P(E) = n(E)/n(S) = 2/2 = 1

Ans: Thus the probability of getting the head or the tail is 1

Note: It is a certain event

f) Getting the head and the tail

Let F be an event of getting the head and the tail

F = {} and n(F) = 0

By definition of probability

P(F) = n(F)/n(S) = 0/2 = 0

Ans: Thus the probability of getting the head and the tail is 0

Note: It is an impossible event

Tossing of Two Coins:

A fair coin is tossed two times is equivalent to two fair coins are tossed.

The sample space is 

S = {HH, HT, TH, TT} 

n(S) = 4

Example – 02:

Two unbiased coins are tossed. Find the probability of getting  OR A fair coin is tossed two times. Find the probability of getting

Solution:

Two unbiased coins are tossed or a fair coin tossed twice.

The sample space for the experiment is

S = {HH, HT, TH, TT}

∴  n(S) = 4

a) Exactly one head

Let A be the event of getting exactly one head

∴ A = {HT, TH}

∴ n(A) = 2

By the definition P(A) = n(A)/n(S) = 2/4 = 1/2

Ans: the probability of getting exactly one head is 1/2

b) at least one head

Let B be the event of getting at least one head i.e. one head or two head

∴ B = {HH, HT, TH}

∴ n(B) = 3

By the definition P(B) = n(B)/n(S) = 3/4

Ans: the probability of getting at least one head is 3/4

c) at the most one head

Let C be the event of getting at most one head i.e. no head or one head

∴ C = {HT, TH, TT}

∴ n(C) = 3

By the definition P(C) = n(C)/n(S) = 3/4

Ans: the probability of getting at most one head is 3/4

d) both head

Let D be the event of getting both head

∴ D = {HH}

∴ n(D) = 1

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

Ans: the probability of getting both head is 1/4

e) no head

Let E be the event of getting no head

∴ E = {TT}

∴ n(E) = 1

By the definition P(E) = n(E)/n(S) = 1/4

Ans: the probability of getting no head is 1/4

f) no head on the first coin

Let F be the event of getting no head on the first coin

∴ F = {TH, TT}

∴ n(F) = 2

By the definition P(F) = n(F)/n(S) = 2/4 = 1/2

Ans: the probability of getting no head on first coin is 1/2

g) no head on the second coin

Let G be the event of getting no head on the second coin

∴ G = {HT, TT}

∴ n(G) = 2

By the definition P(G) = n(G)/n(S) = 2/4 = 1/2

Ans: the probability of getting no head on the second coin is 1/2

h) Head on one coin and tail on the other

Let H be the event of getting head on one coin and tail on the other

∴ H = {HT, TH}

∴ n(H) = 2

By the definition P(H) = n(H)/n(S) = 2/4 = 1/2

Ans: the probability of getting head on one coin and tail on the other is 1/2

i) Head on the first coin and the tail on the other

Let J be the event of getting head on the first coin and the tail on the other

∴ J = {HT}

∴ n(J) = 1

By the definition P(J) = n(J)/n(S) = 1/4

Ans: the probability of getting head on the first coin and the tail on the other is 1/4

Tossing of Three Coins:

A fair coin is tossed three times is equivalent to three fair coins are tossed.

The sample space is 

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

Example – 03:

Three coins are tossed (OR A coin is tossed three times) and the results are recorded. Find the probabilities in the following events

Solution:

Three unbiased coins are tossed or a fair coin tossed thrice.

The sample space for the experiment is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

∴  n(S) = 8

a) getting exactly one head

Let A be the event of getting exactly one head

∴ A = { HTT, THT, TTH}

∴ n(A) = 3

By the definition P(A) = n(A)/n(S) = 3/8

Ans: the probability of getting exactly one head is 3/8

b) getting exactly two heads

Let B be the event of getting exactly two heads

∴ B = {HHT, HTH, THH}

∴ n(B) = 3

By the definition P(B) = n(B)/n(S) = 3/8

Ans: the probability of getting exactly two heads is 3/8

c) getting all heads

Let C be the event of getting all heads

∴ C = {HHH}

∴ n(C) = 1

By the definition P(C) = n(C)/n(S) = 1/8

Ans: the probability of getting all heads is 1/8

d) getting two or more heads (at least two heads):

Let D be the event of getting atleast  two heads i.e. getting two or three heads

∴ D = {HHH, HHT, HTH, THH}

∴ n(D) = 4

By the definition P(D) = n(D)/n(S) = 4/8 = 1/2

Ans: the probability of getting at least two heads is 1/2

e) getting no head:

Let E be the event of getting no head

∴ E = {TTT}

∴ n(E) = 1

By the definition P(E) = n(E)/n(S) = 1/8 = 1/

Ans: the probability of getting no head is 1/8

f) getting at least one head:

Let F be the event of getting atleast one head

p(atleast one head) = 1 – P(no head) = 1- 1/8 = 7/8

Ans: the probability of getting at least one head is 7/8

g) getting atmost one head:

Let G be the event of getting atmost one head i.e. getting no head or one head

G = {HTT, THT, TTH, TTT}

∴ n(G) = 4

By the definition P(G) = n(G)/n(S) = 4/8 = 1/2

Ans: the probability of getting atmost one head is 1/2

h) getting atmost two heads:

Let H be the event of getting atmost two head i.e. getting no head or one head

H = {HHT, HTH, THH, HTT, THT, TTH, TTT}

∴ n(H) = 7

By the definition P(H) = n(H)/n(S) = 7/8

Ans: the probability of getting atmost two heads is 7/8

i) getting head on second toss or second coin:

Let J be the event of getting head on second toss

J = {HHH, HHT, THH, THT}

∴ n(J) = 4

By the definition P(J) = n(J)/n(S) = 4/8 = 1/2

Ans: the probability of getting head on the second toss is 1/2

Tossing of Four Coins:

Example – 04:

Four coins are tossed and the results are recorded. Find the probabilities in the following events

Solution:

four unbiased coins are tossed the number of points in sample space for the experiment is

∴  n(S) = 2⁴ = 16

a) getting exactly one head

Let A be the event of getting exactly one head

∴ A = { HTTT, THTT, TTHT, TTTH}

∴ n(A) = 4

By the definition P(A) = n(A)/n(S) = 4/16

Ans: the probability of getting exactly one head is 1/4

b) getting no head

Let B be the event of getting exactly one head

∴ B = { TTTT}

∴ n(B) = 1

By the definition P(B) = n(B)/n(S) = 1/16

Ans: the probability of getting exactly no head is 1/16

c) getting at least one head

Let C be the event of getting atleast one head

p(atleast one head) = 1 – P(no head) = 1- 1/16 = 15/16

Ans: the probability of getting at least one head is 15/16

In the next article, we shall study some basic problems of probability based on the tossing of a single die.

For More Topics in Probability Click Here

For More Topics in Mathematics Click Here

The post Problems Based on Tossing of Coins appeared first on The Fact Factor.