UPSC NDA Math Model Paper 2

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=

5

21