JEE Main 2025 — Previous Year Question.
Problem
Two biased dice are tossed. Die 1 has 1 on two faces, 2 on two faces, 3 and 4 on the other faces, while die 2 has 2 on two faces, 4 on two faces, and 1 and 3 on the other faces. The probability that, when throwing these dice, we get a sum of 4 or 5 is:
Key insight. With biased dice, each face’s probability out of 6 depends on how many of the die’s 6 physical faces carry that number — once those per-face probabilities are written down, the two dice are still independent, so every combination that sums to 4 or 5 is just a product of two probabilities, summed over all such combinations.
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Approach
Write down the probability of each face value on each die, based on how many of its 6 faces carry that number. List every ordered pair (die 1 value, die 2 value) that sums to 4, and every pair that sums to 5. Since the two dice are independent, multiply the two face-probabilities for each pair, then add up all the contributing pairs.
Solution
Step 1 — Write the face probabilities for each die
Die 1: two faces show 1, two show 2, one shows 3, one shows 4:
$$P_1(1)=\frac26,\ P_1(2)=\frac26,\ P_1(3)=\frac16,\ P_1(4)=\frac16$$
Die 2: two faces show 2, two show 4, one shows 1, one shows 3:
$$P_2(1)=\frac16,\ P_2(2)=\frac26,\ P_2(3)=\frac16,\ P_2(4)=\frac26$$
Step 2 — List the pairs summing to 4
$(1,3)$, $(2,2)$, $(3,1)$ — where the first number is die 1’s value and the second is die 2’s.
$$P(1,3)=P_1(1)P_2(3)=\frac26\cdot\frac16=\frac2{36}$$
$$P(2,2)=P_1(2)P_2(2)=\frac26\cdot\frac26=\frac4{36}$$
$$P(3,1)=P_1(3)P_2(1)=\frac16\cdot\frac16=\frac1{36}$$
$$P(\text{sum}=4) = \frac2{36}+\frac4{36}+\frac1{36}=\frac7{36}$$
Step 3 — List the pairs summing to 5
$(1,4)$, $(2,3)$, $(3,2)$, $(4,1)$:
$$P(1,4)=P_1(1)P_2(4)=\frac26\cdot\frac26=\frac4{36}$$
$$P(2,3)=P_1(2)P_2(3)=\frac26\cdot\frac16=\frac2{36}$$
$$P(3,2)=P_1(3)P_2(2)=\frac16\cdot\frac26=\frac2{36}$$
$$P(4,1)=P_1(4)P_2(1)=\frac16\cdot\frac16=\frac1{36}$$
$$P(\text{sum}=5) = \frac4{36}+\frac2{36}+\frac2{36}+\frac1{36}=\frac9{36}$$
Step 4 — Add both totals
$$P(\text{sum}=4 \text{ or } 5) = \frac7{36}+\frac9{36} = \frac{16}{36} = \frac49$$
Answer
$$\frac49$$
Common mistakes
- Treating the dice as unbiased. With a fair die, each face would have probability $\dfrac16$ regardless of the value — here, values 1 and 2 on die 1 (and 2 and 4 on die 2) are each twice as likely as the other values, since two physical faces carry them.
- Missing a contributing pair. With four ordered pairs summing to 5 alone, it’s easy to list only two or three — systematically going through die 1’s value from smallest to largest and finding the matching die 2 value each time avoids missing one.
- Adding probabilities instead of multiplying for a single pair. Each ordered pair like $(2,2)$ requires both dice to show specific values simultaneously — since the throws are independent, this is a product ($P_1(2)\times P_2(2)$), not a sum.
Practise next
- Using the same two biased dice, find the probability of getting a sum of 6 or 7.
Show answer
$\dfrac5{18}$. Sum of 6: $(2,4),(3,3),(4,2)$, giving $\dfrac4{36}+\dfrac1{36}+\dfrac2{36}=\dfrac7{36}$.
Sum of 7: $(3,4),(4,3)$, giving $\dfrac2{36}+\dfrac1{36}=\dfrac3{36}$.
Total: $\dfrac7{36}+\dfrac3{36}=\dfrac{10}{36}=\dfrac5{18}$.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.