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

(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

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

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Probability for Rolling Two Dice

Solved Probability Problems

Probability for Rolling Three Dice

9th Grade Math

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