If X follows Binomial distribution with parameters n=5, p and P(X=2)=9P(X=3), then p is equal to ……. .

Explanation

`becauseP(X=2)=9cdotP(X=3)` (where, n=5 and q=1-p)
`rArr""^(5)C_(2)p^(2)(1-p)^(3)=9cdot""^(5)C_(3)p^(3)(1-p)^(2)`
`rArr(5!)/(2!3!)p^(2)(1-p)^(3)=9cdot(5!)/(3!2!)p^(3)(1-p)^(2)`
`rArr(p^(2)(1-p)^(3))/(p^(3)(1-p)^(2))=9`
`rArr((1-p)/p)=9rArr9p+p=1`
`thereforep=1/10`