JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
Let X be a binomially distributed random variable with mean 4 and variance 4/3. Then, 54 P(X ≤ 2) is equal to :
- (a) 73/27
- (b) 146/27
- (c) 146/81
- (d) 126/81
Answer and solution
Answer (b) 146/27
Solution
Maths sir · 3 steps
Mean and variance together pin down n and p; after that it is three terms of the binomial.
- Find q, p, n. Mean np = 4 and variance npq = 4/3. Divide the second by the first: q = 1/3, so p = 2/3 and n = 4/(2/3) = 6.
P(X ≤ 2) = C(6, 0)q⁶ + C(6, 1)p q⁵ + C(6, 2)p²q⁴
- Put in numbers. Every term has denominator 3⁶ = 729. The numerators are 1, 6 × 2 = 12 and 15 × 4 = 60. So P(X ≤ 2) = (1 + 12 + 60)/729 = 73/729.
- So. 54 × 73/729 = (73 × 2)/27 = 146/27, which is option (b).
The idea to keep. For a binomial, variance divided by mean gives q straight away. Then n follows from np, and every probability is a sum of C(n, r)pʳqⁿ⁻ʳ terms.
Answer from the NTA final answer key. Solution by PrepWiser.