NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 13: Statistics and Probability - Exercise 13.3

Question 22

Step 1: When two dice are thrown, each die can show any number from 1 to 6. So, the total number of outcomes = \(6 \times 6 = 36\).

Step 2: We want the product of the two numbers to be less than 9. Let's list all such pairs (ordered pairs are written as (first die, second die)):

• When first die = 1 → Products = (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) → all products < 9 → 6 outcomes
• When first die = 2 → (2,1), (2,2), (2,3), (2,4) → products = 2, 4, 6, 8 (all < 9) → 4 outcomes
• When first die = 3 → (3,1), (3,2) → products = 3, 6 (< 9) → 2 outcomes
• When first die = 4 → (4,1), (4,2) → products = 4, 8 (< 9) → 2 outcomes
• When first die = 5 → (5,1) → product = 5 (< 9) → 1 outcome
• When first die = 6 → (6,1) → product = 6 (< 9) → 1 outcome

Step 3: Total favourable outcomes = \(6 + 4 + 2 + 2 + 1 + 1 = 16\).

Step 4: Probability = \[ \dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} = \dfrac{16}{36} = \dfrac{4}{9}. \]

Final Answer: The probability is \(\dfrac{4}{9}\).