A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is , gcd (m, n) = 1, then is equal to [2025]

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Solution

Q.
A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is $\frac{m}{n}$ , gcd (m, n) = 1, then $n^{2} - m^{2}$ is equal to [2025]

1 64

2 80

3 72

4 60

Ans.
(2)

Let us defined the events as

A : Unbiased coin is selected

B : Biased coin in selected

H : Head turns up.

$∴ P (H) = P (A) P (H / A) + P (B) P (H / B)$

$= \frac{19}{20} \times \frac{1}{2} + \frac{1}{20} \times 1 = \frac{21}{40}$

Now, $P (A / H) = \frac{P (A) P (H / A)}{P (H)} = \frac{\frac{19}{20} \times \frac{1}{2}}{\frac{21}{40}} = \frac{19}{21}$

$\Rightarrow \frac{m}{n} = \frac{19}{21}$

$∴ n^{2} - m^{2} = 441 - 361 = 80$ .

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