JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Probability

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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 :

(A)80✓ Correct
(B)60
(C)72
(D)64

Answer: (A)

Solution

By Bayes' theorem,

P (unbiased ∣ head) = \frac{P ( unbiased ) P ( head ∣ unbiased )}{P ( head )}

. Here

P (unbiased) = \frac{19}{20}

with

P (head ∣ unbiased) = \frac{1}{2}

, while

P (biased) = \frac{1}{20}

with

P (head ∣ biased) = 1

. Thus

P (head) = \frac{19}{20} \cdot \frac{1}{2} + \frac{1}{20} = \frac{21}{40}

, and

P (unbiased ∣ head) = \frac{19/40}{21/40} = \frac{19}{21}

. With

m = 19, n = 21

n^{2} - m^{2} = 441 - 361 = 80

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Single correct option

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