Given, if 'X' has a binomial distribution with parameters n = 6, p and P(X = 2) = 12, P(X = 3) = 5 We know that, P(X=r)=nCrPr(1−P)n−r ⇒P(X=2)=6C2P2(1−P)6−2 ⇒12=6C2P2(1−P)4 ⇒12=15P2(1−P)4.......(1) Now, ⇒P(X=3)=6C3P3(1−P)6−3 ⇒5=20P3(1−P)3......(2) By (1) / (2), we get ⇒
12
5
=
15P2(1−P)4
20P3(1−P)3
⇒
12
5
=
3(1−P)
4P
⇒48P=15−15P ⇒63P=15 ⇒P=
5
21
Hence, if 'X' has a binomial distribution with parameters n = 6, p and P(X = 2) = 12, P(X = 3) = 5 then P=