# Tossing a Coin

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### Problems Based on Tossing of Two Unbiased 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}  and n(S) = 4

#### Example – 01:

• 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

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

Therefore the probability of getting exactly one head is 1/2

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

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

• 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

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

Let D be the event of getting no head

∴ D = {TT}

∴ n(D) = 1

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

Therefore the probability of getting no head is 1/4

• no head on the first coin

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

∴ E = {TH, TT}

∴ n(E) = 1

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

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

Problems based on three unbiased 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} and n(S) = 8

#### Example – 02:

• 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

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

Therefore the probability of getting exactly one head is 3/8

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

Therefore the probability of getting exactly two heads is 3/8

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

Therefore the probability of getting all heads is 1/8

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

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

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/

Therefore the probability of getting no head is 1/8

Let F be the event of getting atleast one head

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

Therefore the probability of getting atleast one head is 7/8

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

Therefore the probability of getting atmost one head is 1/2

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

Therefore the probability of getting atmost two heads is 7/8

• 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

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

#### Example – 03:

• 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

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

Therefore the probability of getting exactly one head is 1/4

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

Therefore the probability of getting exactly no head is 1/16