JEE Main 2025 — Probability Question with Solution

Define the Events:

Let $A$ be the event that an unbiased coin is drawn.

Let $B$ be the event that the coin with heads on both sides is drawn.

Let $H$ be the event that a head turns up when the coin is tossed.

State the Given Probabilities:

$P(A)=\frac{19}{20}$ (since there are 19 unbiased coins out of 20 total coins)

$P(B)=\frac{1}{20}$ (since there is 1 coin with two heads out of 20 total coins)

$P(H|A)=\frac{1}{2}$ (probability of getting a head given an unbiased coin is drawn)

$P(H|B)=1$ (probability of getting a head given the coin with two heads is drawn)

Use Bayes' Theorem:

We want to find $P(A|H)$, which is the probability that the drawn coin was unbiased given that a head turned up. Bayes' Theorem states:

$$P(A|H)=\frac{P(H|A)\cdot P(A)}{P(H)}$$

Calculate $P(H)$:

We can find $P(H)$ using the law of total probability:

$$P(H)=P(H|A)\cdot P(A)+P(H|B)\cdot P(B)$$

$$P(H)=\left(\frac{1}{2}\right)\cdot \left(\frac{19}{20}\right)+(1)\cdot \left(\frac{1}{20}\right)=\frac{19}{40}+\frac{1}{20}=\frac{19}{40}+\frac{2}{40}=\frac{21}{40}$$

Apply Bayes' Theorem to find $P(A|H)$:

$$P(A|H)=\frac{P(H|A)\cdot P(A)}{P(H)}=\frac{\left(\frac{1}{2}\right)\cdot \left(\frac{19}{20}\right)}{\frac{21}{40}}=\frac{\frac{19}{40}}{\frac{21}{40}}=\frac{19}{21}$$

Determine $m$ and $n$:

Since $P(A|H)=\frac{m}{n}$ and $\gcd (m,n)=1$, we have $m=19$ and $n=21$.

Calculate $n^2-m^2$:

$$n^2-m^2=21^2-19^2=(21+19)(21-19)=(40)(2)=80$$

Therefore, $n^2-m^2=80$.

So, the correct answer is:

Option B: 80