Since X has a binomial distribution, B(n,p) ∴P(X=2)=nC2(p)2(1−p)n−2 and P(X=3)=nC3(p)3(1−p)n−3 Given P(X=2)=P(X=3) ⇒nC2p2(1−p)n−2=nC3(p)3(1−p)n−3 ⇒
n!
2!(n−2)!
.
p2(1−p)n
(1−p)2
=
n!
3!(n−3)!
.
p3(1−p)n
(1−p)3
⇒
1
n−2
=
1
3
.
p
1−p
⇒3(1−p)=p(n−2) ⇒3−3p=np−2p ⇒np=3−p ⇒E(X)= mean =3−p (∵ mean of B (n, p) = np)