Two fair dice are thrown simultaneously. Find the probability that: (i) the product of the numbers appearing on the two dice is a prime number, and (ii) the numbers appearing on the two dice are different.

hard 5 marks Class 10 · Mathematics · Probability

Answer

(i)

\frac{1}{6}

, (ii)

\frac{5}{6}

The correct answer is (i) $\frac{1}{6}$ , (ii) $\frac{5}{6}$ — Subtract doublets from total outcomes to find outcomes with different numbers: $36 - 6 = 30$ . Thus, $P (different numbers) = \frac{30}{36} = \frac{5}{6}$ .

Step-by-step solution

Find the total number of outcomes when two dice are thrown: $n (S) = 6 \times 6 = 36$ .
For part (i), list outcomes where the product of the two numbers is a prime number ( $2, 3, 5$ ): $(1, 2), (2, 1)$ give product $2$ ; $(1, 3), (3, 1)$ give product $3$ ; $(1, 5), (5, 1)$ give product $5$ .
Count the favorable outcomes for part (i): there are $6$ outcomes. Thus, $P (product is prime) = \frac{6}{36} = \frac{1}{6}$ .
For part (ii), list the outcomes where both numbers are the same (doublets): $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ , which gives $6$ outcomes.
Subtract doublets from total outcomes to find outcomes with different numbers: $36 - 6 = 30$ . Thus, $P (different numbers) = \frac{30}{36} = \frac{5}{6}$ .

Common mistake: Forgetting that

(1, 2)

and

(2, 1)

are distinct outcomes when rolling two dice.

Concept tested: Probability with rolling two dice

Published 24 August 2026 · How this solution was prepared and checked

Two fair dice are thrown simultaneously. Find the probability that: (i) the product of the numbers appearing on the two dice is a prime number, and (ii) the numbers appearing on the two dice are different.

Step-by-step solution

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