An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is :

Solution

Maths sir · 4 steps

Odd and even each have probability one half, so the condition becomes C(n, 2) = C(n, 3).

1. Set up. Let the die be rolled n times. Odd and even each have probability 1/2, so the number of odd results follows a binomial distribution with p = q = 1/2.
2. Use the condition. Getting an even number 3 times is the same as getting an odd number n - 3 times. Both probabilities share the factor (1/2)ⁿ, so only the binomial coefficients need to match.

   C(n, 2) = C(n, 3) ⇒ 2 + 3 = n ⇒ n = 5

3. Odd number of times. An odd number shows up 1, 3 or 5 times. The probability is (C(5, 1) + C(5, 3) + C(5, 5))/2⁵ = (5 + 10 + 1)/32 = 16/32.
4. So. The probability is 1/2, which is option (c).

The idea to keep. When p = q = 1/2, every outcome count has the same factor 1/2ⁿ, so compare only the C(n, r) terms. The odd-index coefficients always add up to 2ⁿ⁻¹.

Answer from the NTA final answer key. Solution by PrepWiser.