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

Answer: 37

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 $p+q=4+33=37$