JEE Main 2025 Apr 7 Shift 2, Mathematics Q5: Total Probability Theorem And Bayes' Theorem

JEE Main2025Apr 7, Shift 2Mathematics

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 $\frac{m}{n}$ , $g cd (m, n) = 1$ , then $n^{2} - m^{2}$ is equal to :

A
80
B
60
C
72
D
64

Solution

By the total probability rule, $P (H) = P (unbiased) \cdot P (H ∣ unbiased) + P (biased) \cdot P (H ∣ biased) = \frac{19}{20} \times \frac{1}{2} + \frac{1}{20} \times 1 = \frac{19}{40} + \frac{2}{40} = \frac{21}{40}$ .

By Bayes' theorem, $P (unbiased ∣ H) = \frac{P ( unbiased ) P ( H ∣ unbiased )}{P ( H )} = \frac{19/40}{21/40} = \frac{19}{21}$ .

Since $g cd (19, 21) = 1$ , $m = 19, n = 21$ , so $n^{2} - m^{2} = 441 - 361 = 80$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q5: Total Probability Theorem And Bayes' Theorem

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