Probability of Tossing Two Coins

Here we will learn how to find the probability of tossing two coins.

Let us take the experiment of tossing two coins simultaneously:

When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.

Therefore, total numbers of outcome are 2² = 4

The above explanation will help us to solve the problems on finding the probability of tossing two coins.

Worked-out problems on probability involving tossing or flipping two coins:

1. Two different coins are tossed randomly. Find the probability of:

(i) getting two heads

(ii) getting two tails

(iii) getting one tail

(iv) getting no head

(v) getting no tail

(vi) getting at least 1 head

(vii) getting at least 1 tail

(viii) getting atmost 1 tail

(ix) getting 1 head and 1 tail

Solution:

When two different coins are tossed randomly, the sample space is given by

S = {HH, HT, TH, TT}

Therefore, n(S) = 4.

(i) getting two heads:

Let E₁ = event of getting 2 heads. Then,
E₁ = {HH} and, therefore, n(E₁) = 1.
Therefore, P(getting 2 heads) = P(E₁) = n(E₁)/n(S) = 1/4.

(ii) getting two tails:

Let E₂ = event of getting 2 tails. Then,
E₂ = {TT} and, therefore, n(E₂) = 1.
Therefore, P(getting 2 tails) = P(E₂) = n(E₂)/n(S) = 1/4.

(iii) getting one tail:

Let E₃ = event of getting 1 tail. Then,
E₃ = {TH, HT} and, therefore, n(E₃) = 2.
Therefore, P(getting 1 tail) = P(E₃) = n(E₃)/n(S) = 2/4 = 1/2

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(iv) getting no head:

Let E₄ = event of getting no head. Then,
E₄ = {TT} and, therefore, n(E₄) = 1.
Therefore, P(getting no head) = P(E₄) = n(E₄)/n(S) = ¼.

(v) getting no tail:

Let E₅ = event of getting no tail. Then,
E₅ = {HH} and, therefore, n(E₅) = 1.
Therefore, P(getting no tail) = P(E₅) = n(E₅)/n(S) = ¼.

(vi) getting at least 1 head:

Let E₆ = event of getting at least 1 head. Then,
E₆ = {HT, TH, HH} and, therefore, n(E₆) = 3.
Therefore, P(getting at least 1 head) = P(E₆) = n(E₆)/n(S) = ¾.

(vii) getting at least 1 tail:

Let E₇ = event of getting at least 1 tail. Then,
E₇ = {TH, HT, TT} and, therefore, n(E₇) = 3.
Therefore, P(getting at least 1 tail) = P(E₂) = n(E₂)/n(S) = ¾.

(viii) getting atmost 1 tail:

Let E₈ = event of getting atmost 1 tail. Then,
E₈ = {TH, HT, HH} and, therefore, n(E₈) = 3.
Therefore, P(getting atmost 1 tail) = P(E₈) = n(E₈)/n(S) = ¾.

(ix) getting 1 head and 1 tail:

Let E₉ = event of getting 1 head and 1 tail. Then,
E₉ = {HT, TH } and, therefore, n(E₉) = 2.
Therefore, P(getting 1 head and 1 tail) = P(E₉) = n(E₉)/n(S)= 2/4 = 1/2.

The solved examples involving probability of tossing two coins will help us to practice different questions provided in the sheets for flipping 2 coins.

Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability