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 $\displaystyle P(X=4)=0.09724$ we have:
$\displaystyle P(X=4)=\dfrac{7!}{4!3!}p^4(1-p)^3=0.09724$
Now you need to find $\displaystyle p<0.5$ 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 $\displaystyle p=0.3$)
Now you only need to evaluate $\displaystyle P(X=2)=b(2;7,0.3)$