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A bag contains $(2 n + 1)$ coins. It is known that n of these coins have a head on both sides, whereas the remaining $(n + 1)$ coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is $\frac{31}{42},$ then $n$ is equal to

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Detailed Solution

$E_{1} :$ a biased coin is picked up;

$E_{2} :$ a fair coin is picked up.

$A :$ the toss results in a head.

We have

$\begin{array}{l} P (E_{1}) = \frac{n}{2 n + 1}, P (E_{2}) = \frac{n + 1}{2 n + 1} \\ P (A ∣ E_{1}) = 1, P (A ∣ E_{2}) = \frac{1}{2} \end{array}$

Now, use the total probability

$P (A) = P (E_{1}) P (A ∣ E_{1}) + P (E_{2}) P (A ∣ E_{2}) \Rightarrow \frac{31}{42} = (\frac{n}{2 n + 1}) 1 + (\frac{n + 1}{2 n + 1}) \frac{1}{2} \Rightarrow 21 (3 n + 1) = 31 (2 n + 1) \Rightarrow n = 10$

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