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A box contains N coins, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair?

A box contains N coins, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Hello Student,
Please find the answer to your question
Let E1 be the event that the coin drawn is fair and E2 be the event that the coin drawn is biased.
∴ P (E1) = m/N and P (E2) = N – m/N
A is the event that on tossing the coin head appears first and then appears tail.
∴ P (A) = P (E1 ∩ A) + P (E2 ∩A)
= P (E1) P (A| E1) + P (E2) P (A| E2)
= m/n (1/2)2 + (N – m/N) (2/3) (1/3) . . . . . . . . . . . . . . . . . . . . . . (1)
We have to find the probability that A has happened because of E1
∴ P (E1| A) = P (E1| ∩ A)/P (A)
= \frac{\frac{m}{n}(\frac{1}{2})^2}{\frac{m}{n}(\frac{1}{2})^2 + \frac{N-m}{N}(\frac{2}{3})(\frac{1}{3})}
= \frac{m/4}{m/4+\frac{2(N-m)}{9}}= 9m/m + 8N

Thanks
Aditi Chauhan
askIITians Faculty

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