Mathematics · Probability

JEE Main 2025 — 7 April, Evening Shift — Question 32

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

Option A:
64
Option B:
72
Option C:
80
Correct
Option D:
60

Answer: C

Step-by-step solution

Let $U$ : let unbiased coin is drawn

$H$ : head turns up

$\Rightarrow P(U / H)=\frac{P(U) \cdot P(H / U)}{P(U) P(H / U)+P(\bar{U}) \cdot P(H / \bar{U})}$

\begin{aligned} & =\frac{\frac{19}{20} \cdot \frac{1}{2}}{\frac{19}{20} \cdot \frac{1}{2}+\frac{1}{20} \cdot\left(\frac{2}{2}\right)}=\left(\frac{19}{21}\right)=\frac{m}{n} \\& n^{2}-m^{2}=21^{2}-19^{2}=(21+19)(21-19)=80 \end{aligned}

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Exam: JEE Main 2025
Paper: 7 April, Evening Shift
Subject: Mathematics
Chapter: Probability
Topic: Total Probability and Baye's Theorem

Let a random variable X take values 0,1,2,3 with P(X=0)=P(X=1)=p, P(X=2)= P(X=3)=q and E (X^2 )=2 E(X) . Then the value of 8 p-1 is :

Answer: C

Step-by-step solution

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