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JEE Main 2019 · 12 April · Shift I

JEE Main 2019 Mathematics question on Probability

Let a random variable XX have a binomial distribution with mean 8 and variance 4. If P(X≤2)=k216P(X \leq 2) = \dfrac{k}{2^{16}}, then kk is equal to
  1. A.
    17
  2. B.
    121
  3. C.
    1
  4. D.
    137
Show answer and solution

Correct option: D

Solution

Mean = np = 8, Variance = npq = 4, so q = 1/2, p = 1/2, n = 16.

P(X ≤ 2) = P(0) + P(1) + P(2)
= (1/2)16 · [C(16,0) + C(16,1) + C(16,2)]
= (1 + 16 + 120) / 216
= 137 / 216

So k = 137.

[JEE Main 2019 · 12 April · Shift I]

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