JEE Mathematics 2025 Question with Solution

Given: One die has probabilities $P(D_1=1)=\frac{2}{6},\; P(D_1=2)=\frac{2}{6},\; P(D_1=3)=\frac{1}{6},\; P(D_1=4)=\frac{1}{6}$. The other die has probabilities $P(D_2=1)=\frac{1}{6},\; P(D_2=2)=\frac{2}{6},\; P(D_2=3)=\frac{2}{6},\; P(D_2=4)=\frac{1}{6}$.

Find: The probability that the sum is $4$ or $5$.

Total possible face outcomes are

$$6 \times 6 = 36$$

For sum $4$, the possible ordered pairs are $(1,3), (2,2), (3,1)$. Their probabilities are

$$\frac{2}{6}\times\frac{2}{6}=\frac{4}{36}, \qquad \frac{2}{6}\times\frac{2}{6}=\frac{4}{36}, \qquad \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$

So,

$$P(\text{sum}=4)=\frac{4}{36}+\frac{4}{36}+\frac{1}{36}=\frac{9}{36}=\frac{1}{4}$$

For sum $5$, the possible ordered pairs are $(1,4), (2,3), (3,2), (4,1)$. Their probabilities are

$$\frac{2}{6}\times\frac{1}{6}=\frac{2}{36}, \qquad \frac{2}{6}\times\frac{2}{6}=\frac{4}{36}, \qquad \frac{1}{6}\times\frac{2}{6}=\frac{2}{36}, \qquad \frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$

So,

$$P(\text{sum}=5)=\frac{2}{36}+\frac{4}{36}+\frac{2}{36}+\frac{1}{36}=\frac{9}{36}=\frac{1}{4}$$

Therefore,

$$P(\text{sum}=4 \text{ or } 5)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

Hence, the correct option is A.

Given: The repeated face markings create non-uniform probabilities for the two dice.

Find: The probability that the sum is $4$ or $5$.

Instead of listing all $36$ face outcomes one by one, use the value probabilities directly.

For the first die:

$$P(1)=\frac{1}{3}, \quad P(2)=\frac{1}{3}, \quad P(3)=\frac{1}{6}, \quad P(4)=\frac{1}{6}$$

For the second die:

$$P(1)=\frac{1}{6}, \quad P(2)=\frac{1}{3}, \quad P(3)=\frac{1}{3}, \quad P(4)=\frac{1}{6}$$

Now add probabilities of ordered pairs giving the required sums. For sum $4$:

$$P(4)=P(1,3)+P(2,2)+P(3,1)=\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{4}$$

For sum $5$:

$$P(5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=\frac{1}{3}\cdot\frac{1}{6}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{4}$$

Therefore,

$$P(4 \text{ or } 5)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

This works because the events "sum is $4$" and "sum is $5$" are mutually exclusive. Therefore, the correct option is A.