JEE Main 2026 Apr 6 Shift 1, Mathematics Q10: Total Probability Theorem And Bayes' Theorem

JEE Main2026Apr 6, Shift 1Mathematics

A bag contains $(N + 1)$ coins $-$ $N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$ , then $N$ is equal to:

A
$5$
B
$7$
C
$8$
D
$9$

Solution

By total probability,

$P (H) = \frac{N}{N + 1} \cdot \frac{1}{2} + \frac{1}{N + 1} \cdot 1 = \frac{N + 2}{2 ( N + 1 )}$ .

Set equal to $\frac{9}{16}$ : $16 (N + 2) = 18 (N + 1) \Rightarrow 16 N + 32 = 18 N + 18 \Rightarrow N = 7$ .

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JEE Main 2026 Apr 6 Shift 1, Mathematics Q10: Total Probability Theorem And Bayes' Theorem

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