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

Question 20

Step 1: When two dice are thrown, each die has 6 faces (numbers 1 to 6). So, total possible outcomes = \(6 \times 6 = 36\).

(i) Probability that the sum is 7:

• We need all pairs where the two dice add up to 7.
• Possible pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
• Number of favourable outcomes = 6.
• Probability = \(\tfrac{6}{36} = \tfrac{1}{6}\).

(ii) Probability that the sum is a prime number:

• Prime numbers between 2 and 12 (possible sums): 2, 3, 5, 7, 11.
• Count outcomes for each prime sum:
• Sum = 2 → (1,1) → 1 way
• Sum = 3 → (1,2), (2,1) → 2 ways
• Sum = 5 → (1,4), (2,3), (3,2), (4,1) → 4 ways
• Sum = 7 → (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways
• Sum = 11 → (5,6), (6,5) → 2 ways
• Total favourable outcomes = \(1+2+4+6+2 = 15\).
• Probability = \(\tfrac{15}{36} = \tfrac{5}{12}\).

(iii) Probability that the sum is 1:

• Smallest sum of two dice = 1+1 = 2.
• So, getting sum = 1 is impossible.
• Favourable outcomes = 0.
• Probability = \(\tfrac{0}{36} = 0\).