Question

There are two coins on a table. When both are flipped, one coin lands on heads with probability 0.5 while the other lands on heads with probability 0.6. A coin is randomly selected from the table and flipped. (a) what is the probability it lands on heads? (b) given that it lands on tails, what is the conditional probability that it was the fair coin (that is, the one equally likely to land heads or tails)?

Answer 1

The probability it lands on head is 0.55 and the conditional probability that it was the fair coin is $\frac{5}{11}$

Explanation:

$E_{1}$: Event of selecting a coin in which probability of head is 0.5.

$E_{2}$: Event of selecting a coin in which probability of head is 0.6.

A: Coin lands on a head

(a) Required probability: P(A)

$P(A)=P(E_{1}).P(A/E_{1})+P(E_{2}).P(A/E_{2})$ ...(1)

Here, $P(E_{1})=\frac{1}{2}$ , $P(E_{2})=\frac{1}{2}$, $P(A/E_{1})=0.5$, and $P(A/E_{2})=0.6$

On substituting the values in (1) we have,

$P(A)=0.5\times\frac{1}{2}+0.6\times\frac{1}{2}$

$P(A)=\frac{1}{2}(0.5+0.6)$

$P(A)=\frac{1}{2}(1.1)$

$P(A)=\frac{11}{20}$

$P(A)=0.55$

(b) Required probability: $P(E_{1}/A)$

$P(E_{1}/A)=\frac{P(E_{1}).P(A/E_{1})}{P(E_{1}).P(A/E_{1})+P(E_{2}).P(A/E_{2})}$ ...(2)

On substituting the values in (2) we have,

$P(E_{1}/A)=\frac{\frac{1}{2}\times 0.5}{\frac{1}{2}\times 0.5+\frac{1}{2}\times 0.6}$

$P(E_{1}/A)=\frac{5}{11}$

Learn more:

Conditional probability

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