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 :

P(x=n)=nCx(1−P)n−xPx
nC2(1−P)n−2P2=nC3(1−P)n−3P3
nC2=nC3P/(1−P)
(n−1)/2 = (n−1)(n−2)/6 * P/(1−P)
1/2 = (n−2)/6 * P/(1−P)
3(1−P) = (n−2)P
3−3P = (n−2)P
3 = (n+1)P
P = 3/(n+1)
E(x) = nP
E(x) = n*3/(n+1)
If n is large then np ≈ 3
Given that P(X=2) = P(X=3)
Then nC2 p^2 (1-p)^(n-2) = nC3 p^3 (1-p)^(n-3)
(n(n-1)/2) p^2 (1-p)^(n-2) = (n(n-1)(n-2)/6) p^3 (1-p)^(n-3)
3(1-p) = (n-2)p
3 - 3p = np - 2p
3 = np + p
3 = p(n+1)
p = 3/(n+1)
E(X) = np = n * 3/(n+1)
If n is large, then E(X) ≈ 3
Let's use the given condition P(X=2) = P(X=3):
(n choose 2) * p^2 * (1-p)^(n-2) = (n choose 3) * p^3 * (1-p)^(n-3)
[n(n-1)/2] * p^2 * (1-p)^(n-2) = [n(n-1)(n-2)/6] * p^3 * (1-p)^(n-3)
3(1-p) = (n-2)p
3 - 3p = np - 2p
3 = np + p
3 = p(n+1)
p = 3/(n+1)
E(X) = np = n * 3/(n+1) ≈ 3 when n is large.