Question

There are 6 red balls and 6 green balls in a bag. Five balls are drawn out at random and placed in a red box. The remaining seven balls are put in a green box. If the probability that the number of red balls in the green box plus the number of green balls in the red box is not a prime number, is $\frac{p}{q}$ where $p$ and $q$ are relatively prime, then find the value of $(p+q)$

Answer 1

Let $E$ is event as desired then
$P ( E )=\frac{{ }^6 C _0 \cdot{ }^6 C _5+{ }^6 C _4 \cdot{ }^6 C _1}{{ }^{12} C _5}=\frac{{ }^6 C _1+{ }^6 C _4 \cdot{ }^6 C _1}{11 \cdot 9 \cdot 8}=\frac{6+90}{11 \cdot 9 \cdot 8}=\frac{96}{11 \cdot 9 \cdot 8}=\frac{4}{33}$
hence $p + q = 4 + 33 = 37 $