A probability problem by Shanthanu Rai

Let $p(G_{k})$ be the probability that the bag has $k$ green balls. Then $p(G_{k}) = ak^{2}$ for some constant $a$ , for which we require that

$\displaystyle\sum_{k=0}^{15} p(G_{k}) = 1 \Longrightarrow \sum_{k=0}^{15} ak^{2} = a*\dfrac{15(15 + 1)(2*15 + 1)}{6} = 1 \Longrightarrow a = \dfrac{1}{1240}$ ,

where the formula $\displaystyle\sum_{k=0}^{n} k^{2} = \dfrac{n(n + 1)(2n + 1)}{6}$ was used. Thus $p(G_{k}) = \dfrac{k^{2}}{1240}$ .

Next, to find $P(A)$ , we first note that if event $G_{k}$ occurs, the probability that the randomly chosen ball is red is $\dfrac{15 - k}{15} = 1 - \dfrac{k}{15}$ . We will denote this conditional probability as $P(A|G_{k})$ . So then $P(A)$ can be calculated as

$P(A) = \displaystyle\sum_{k=0}^{15} P(A|G_{k})p(G_{k}) = \sum_{k=0}^{15} \left(\left(1 - \dfrac{k}{15}\right)\dfrac{k^{2}}{1240}\right) =$

$\displaystyle\sum_{k=0}^{15} \dfrac{k^{2}}{1240} - \sum_{k=0}^{15} \dfrac{k^{3}}{18600} = 1 - \dfrac{1}{18600}\left(\dfrac{15(15 + 1)}{2}\right)^{2} = 1 - \dfrac{24}{31} = \dfrac{7}{31}$ ,

where the formula $\displaystyle\sum_{k=0}^{n} k^{3} = \left(\dfrac{n(n + 1)}{2}\right)^{2}$ was used. Thus $q - p = 31 - 7 = \boxed{24}$ .