Mathematics · Probability

JEE Main 2026 — 6 April, Morning Shift — Question 30

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:

Option A:
$5$
Option B:
$7$
Correct
Option C:
$8$
Option D:
$9$

Answer: B

Step-by-step solution

Total $(\mathrm{N}+1)$ coins $\mathrm{N} \Rightarrow$ fair coins $1 \Rightarrow$ Head on both side $\mathrm{P}\left(1^{+}\right)=\left(\frac{\mathrm{N}}{\mathrm{N}+1}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{\mathrm{~N}+1}\right) 1$ $=\frac{\mathrm{N}+2}{2(\mathrm{~N}+1)}=\frac{9}{16}$ $8 \mathrm{~N}+16=9 \mathrm{~N}+9$ $\Rightarrow \mathrm{N}=7$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Mathematics
Chapter: Probability
Topic: Conditional Probability and Multiplication Theorem