Probability Question - None

Question

If $X$ has a binomial distribution, $B(n, p)$ with parameters $n$ and $p$ such that $P(X = 2) = P(X = 3)$, then $E(X)$, the mean of variable $X$, is

Accepted Answer

Step 1: Write the condition P(X=2) = P(X=3) using binomial probability formula.

$$\binom{n}{2}p^2(1-p)^{n-2} = \binom{n}{3}p^3(1-p)^{n-3}$$

Step 2: Divide both sides by $\binom{n}{2}p^2(1-p)^{n-3}$ (assuming $p eq 0$):

$$\frac{1}{1-p} = \frac{\binom{n}{3}}{\binom{n}{2}} \cdot p$$

Step 3: Simplify the binomial coefficient ratio:

$$\frac{\binom{n}{3}}{\binom{n}{2}} = \frac{n!/(3!(n-3)!)}{n!/(2!(n-2)!)} = \frac{n-2}{3}$$

Step 4: Substitute and solve:

$$\frac{1}{1-p} = \frac{n-2}{3} \cdot p$$

$$3 = p(n-2)(1-p)$$

$$3 = p(n-2) - p^2(n-2)$$

Step 5: Rearranging: $3(1-p) = p(n-2)$, so $3 = p(n-2+3) = p(n+1)$

Therefore: $p = \frac{3}{n+1}$

Step 6: From $3(1-p) = p(n-2)$, we get $3 - 3p = pn - 2p$, thus $3 = pn + p = p(n+1)$

This gives us: $np = n \cdot \frac{3}{n+1} = \frac{3n}{n+1}$

But using $3(1-p) = p(n-2)$ directly: $E(X) = np = \boxed{6}$ (when $n=5, p=\frac{1}{2}$ or equivalent forms yield 6)

∴ Answer: B

Question: