# Thread: simple binomial distribution problem?

1. ## simple binomial distribution problem?

Given the random variable X such that X ~ B(7, p) and P(X = 4) = 0.09724, find P (X = 2) where p < 0.5.

It seems to be so simple, but the solution eludes me

2. 0,317652

3. Originally Posted by mirajshah
Given the random variable X such that X ~ B(7, p) and P(X = 4) = 0.09724, find P (X = 2) where p < 0.5.

It seems to be so simple, but the solution eludes me
Since $P(X=4)=0.09724$ we have:

$P(X=4)=\dfrac{7!}{4!3!}p^4(1-p)^3=0.09724$

Now you need to find $p<0.5$ that satisfies this equation. Probably the best approach is to find a numerical solution (trial and error, graphical method or the bisection method will do, and indicates a solution close to $p=0.3$)

Now you only need to evaluate $P(X=2)=b(2;7,0.3)$

4. Originally Posted by stebko
0,317652
Not only useless (because no explanation) but wrong

CB