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A bag contains $(2 n + 1)$ coins. It is known that n of these coins have head on both the sides, where as 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 find the value of n.

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answer is 10.

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Detailed Solution

Let A: The coin is fair
B: The coin biased
Now $P (H) = P (A) P (\frac{H}{A}) + P (B) P (\frac{H}{B}),$ (using total law of probability)

$\Rightarrow [\frac{n + 1}{2 n + 1}] \times \frac{1}{2} + [\frac{n}{2 n + 1}] \times 1 = \frac{31}{42}$ (given)

$\frac{3 n + 1}{2 (2 n + 1)} = \frac{31}{42} \Rightarrow \frac{3 n + 1}{2 n + 1} = \frac{31}{21}$

$\Rightarrow n = 10$

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