A fair coin is tossed $n$ times. Let $X$ denote the number of times head occurs. If $P(X = 4)$, $P(X = 5)$ and $P(X = 6)$ are in arithmetic progression, then the value of $n$ can be:
Step-by-Step Solution
Key Concept: For a binomial distribution with n trials and probability 1/2, P(X=k) = C(n,k)/2^n. The condition that P(X=4), P(X=5), P(X=6) are in AP translates to 2·C(n,5) = C(n,4) + C(n,6), which reduces to a quadratic equation in n after using the recurrence relation C(n,k)/C(n,k-1) = (n-k+1)/k.
Given that $^nC_4$, $^nC_5$, and $^nC_6$ are in arithmetic progression, we use $2 · ^nC_5 = ^nC_4 + ^nC_6$. Expanding binomial coefficients and simplifying: $\frac{2(n-4)}{5} = 1 + \frac{(n-4)(n-5)}{5·6}$ leads to $n^2 - 21n + 98 = 0$, giving $(n-7)(n-14) = 0$. Thus $n = 7$ or $n = 14$.
Correct Answer: 1,4