Sample 2026-27 Class 12 Maths Q17 Solution: At least one 4, given the sum on two dice is a multiple of 5

The question

A die is thrown twice. What is the probability that at least one number is 4, given the sum of the numbers appearing is a multiple of 5?

(A) $\frac{3}{5}$

(B) $\frac{4}{5}$

(C) $\frac{3}{7}$

(D) $\frac{4}{7}$

Exam answer

Let E: sum is a multiple of 5 and F: at least one number is 4.

There are 36 equally likely outcomes. The sum lies between 2 and 12, so a multiple of 5 means a sum of 5 or 10:

\begin{aligned} E = \{&(1, 4), (2, 3), (3, 2), \\ &(4, 1), (4, 6), (5, 5), (6, 4)\} \end{aligned}

Why?

Sum 5: $(1, 4), (2, 3), (3, 2), (4, 1)$ ⁠. Sum 10: $(4, 6), (5, 5), (6, 4)$ ⁠. So $n(E) = 7$ ⁠.

Order matters: $(1, 4)$ and $(4, 1)$ are different outcomes of the two throws.

The outcomes of $E$ with at least one 4 are

\begin{aligned} E \cap F = \{&(1, 4), (4, 1), \\ &(4, 6), (6, 4)\} \end{aligned}

Why?

$(2, 3)$ ⁠, $(3, 2)$ and $(5, 5)$ have no 4, so they are left out. So $n(E \cap F) = 4$ ⁠.

\begin{aligned} P(F|E) &= \frac{P(E \cap F)}{P(E)} = \frac{4/36}{7/36} \\ &= \frac{n(E \cap F)}{n(E)} = \frac{4}{7} \end{aligned}

Why?

Conditional probability, for $P(E) \ne 0$ ⁠:

P(F|E) = \frac{P(E \cap F)}{P(E)}

Dividing by a fraction is multiplying by it upside down:

\begin{aligned} \frac{4/36}{7/36} &= \frac{4}{36} \times \frac{36}{7} \\ &= \frac{4 \times 36}{36 \times 7} = \frac{4}{7} \end{aligned}

The 36 cancels, so we simply count inside $E$ ⁠. Watch out: dividing by 36 instead gives $\frac{4}{36}$ ⁠, which is $P(E \cap F)$ ⁠, not the answer.

Answer: (D) $\dfrac{4}{7}$

Where this comes from

Chapter 13, Probability: Conditional probability.

Question type: Dice, coins and cards

How to spot it: click to see

Type 2. Dice, coins and cards: shrink the sample space

Spot it: equally likely outcomes, and the words "given that" or "it is known that".

Here: "Given the sum is a multiple of 5" shrinks the 36 outcomes to the 7 with sum 5 or 10. Four of them contain a 4, so the answer is $\frac47$ ⁠.

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Q17: At least one 4, given the sum on two dice is a multiple of 5