Multiple choice

A fair coin is tossed n times and let X denote the number of heads obtained. If P(X = 4), P(X = 5) and P(X = 6) are in A.P., then n is equal to

  1. 7

  2. 14

  3. 7 or 14

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

P(X=k) = C(n,k) * (1/2)^n. If P(4), P(5), P(6) are in AP, then 2*P(5) = P(4) + P(6). 2*C(n,5) = C(n,4) + C(n,6). Dividing by C(n,5) leads to 2 = (n-4)/5 + 5/(n-5). Solving this quadratic gives n = 7 or n = 14.

AI explanation

For a binomial distribution, the probabilities are P(X=r) = nCr × (1/2)^n. Since P(X=4), P(X=5), and P(X=6) are in arithmetic progression, the middle term must be the average of the other two, so 2 × nC5 = nC4 + nC6. Expanding these combinations yields 2 × n! / (5! (n-5)!) = n! / (4! (n-4)!) + n! / (6! (n-6)!). Dividing by n! / (6! (n-4)!) simplifies the equation to 12(n-4) = 30 + (n-4)(n-5), resulting in the quadratic equation n^2 - 21n + 98 = 0. Solving this gives n = 7 or n = 14.

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