CBSE Previous Year Paper

2017 Class 10 Math Set 3 Q23. Probability

Question: Two different dice are thrown together. Find the probability that the numbers obtained have
  1. even sum, and
  2. even product

Video Explanation

Explanatory Answer

(i) The sum of the numbers is even.

Sum of two numbers is even if
(a) both numbers are even (or)
(b) both numbers are odd.

(a) Both even: 3 possibilities for each die to turn out even i.e., 2, 4, and 6.
3 possibilities for one die and 3 for the other i.e., 3 × 3 = 9 outcomes where both are even.
The outcomes are (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6)

(b) Both odd: 3 possibilities for each die to turn out odd i.e., 1, 3, and 5.
3 possibilities for one die and 3 for the other i.e., 3 × 3 = 9 outcomes where both are odd.
The outcomes are (1, 1)(1, 3)(1, 5)(3, 1)(3, 3)(3, 5)(5, 1)(5, 3)(5, 5)

Total outcomes when sum is even = 9 + 9 = 18
Total outcomes when 2 different dice are thrown = 6 × 6 = 36

Probability that sum is even = \\frac{\text{number of outcomes when sum is even}}{\text{Total number of outcomes}}) = \\frac{18}{36}) = \\frac{1}{2})

(ii) Product is even when

(a) one of the dice turns out even and the other turns out odd.
(b) both are even.

(a) One is even and the other is odd.
3 even numbers. 3 odd numbers. First one odd and second even; first even and second odd = 3 × 3 × 2 = 18 outcomes
The outcomes are
(1, 2)(1, 4)(1, 6)(3, 2)(3, 4)(3, 6)(5, 2)(5, 4)(5, 6)
(2, 1)(2, 3)(2, 5)(4, 1)(4, 3)(4, 5)(6, 1)(6, 3)(6, 5)

(b) Both are even.
3 possibilities for the first and 3 for the second = 3 × 3 = 9 outcomes.
The outcomes are (2, 2)(2, 4)(2, 6)(4, 2)(4, 4)(4, 6)(6, 2)(6, 4)(6, 6) = 9 outcomes

Total outcomes when product is even = 18 + 9 = 27
Total outcomes when two different dice are thrown = 6 × 6 = 36

Probability that product is even = \\frac{\text{number of outcomes when product is even}}{\text{Total number of outcomes}}) = \\frac{27}{36}) = \\frac{3}{4})

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