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

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answer is A.

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Detailed Solution

Here,
Total coins =2n+1
Here, when probability for choosing biased coin,

P (B) = \frac{n}{2 n + 1}

For choosing unbiased coin

P (UB) = \frac{n + 1}{2 n + 1}

Now, for head,

P (Head) = \frac{n}{2 n + 1} \times 1 + \frac{n + 1}{2 n + 1} \times \frac{1}{2}

= \frac{2 n + (n + 1)}{2 (2 n + 1)}

= \frac{3 n + 1}{2 (2 n + 1)}

So, we have,

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

(3 n + 1) 42 = (2 n + 1) 62

126 n + 42 = 124 n + 62

2 n = 62 - 42

2 n = 20

\Rightarrow n = 10

Correct option is 1.

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