Baye's Theorem

The probability of drawing one white ball and one green ball from the first urn is $\frac{1}{5}$ .

The probability of drawing one white ball and one green ball from the second urn is $\frac{1}{3}$ .

The probability of drawing one white ball and one green ball from the third urn is $\frac{2}{11}$ .

Therefore, the probability that the third urn was chosen is $\frac{\frac{2}{11}}{\frac{1}{5} + \frac{1}{3} + \frac{2}{11}} = \frac{15}{59}.$ So the answer is $15 + 59 = 74$ .

Hey! guys, I am pretty much sure that this is the answer, but still, if you are getting a different answer (especially 15/59) then feel free to comment in the section below with a valid explanation, I will certainly edit the answer Thanks for the patience till then. And Here's the Solution: $$ By Conditional Probability we have: $P(x\mid B)=\frac{P(x\cap B)}{P(B)}-----------\mid (1)$ Now, if the event $B$ is made of dependent mutually exclusive and exhaustive set of elementary events say ${A_1,A_2,...,A_n}$ then by $1$ we get $P(A_i\mid B)=\frac{P(A_i\cap B)}{P(B)}$ for all $i \in {1,2,...,n}$ . Now we have the following expressions $P(B)=P(A_1 \cup A_2 \cup ...\cup A_n)$ $=P((A_1 \cap B) \cup (A_2 \cap B) ... \cup (A_n \cap B))=\sum P(B \mid A_i).P(A_i)--------------\mid(2)$ and $P(A_i\cap B)=P(B \mid A_i)P(A_i)------------\mid (3)$ Now Use (2) and (3) in (1) and replace $x$ by $A_i$ to get $P(A_i\mid B)=\frac{P(B \mid A_i)P(A_i)}{\sum_{j=1}^{n} P(B \mid A_j)P(A_j)}$ which is the Baye's Theorem; $$ Now in our case assuming $$ $A_1$ : the event of drawing $1W,1R$ from $1st$ Urn; $$ $A_2$ : the event of drawing $1W,1R$ from $2nd$ Urn; $$ $A_3$ : the event of drawing $1W,1R$ from $3rd$ Urn; and $B$ the event of drawing $1W,1R$ So by Baye's Theorem we get,

$P(A_3 \mid B)=\frac{P(B \mid A_3)P(A_3)}{\sum_{i=1}^{3}P(B \mid A_i)P(A_i)}=\frac{(1/4)(1/3)(1/4)}{\frac{1}{3}((1/6)(1/2)+(1/2)(1/4)+(1/3)(1/4))}=\frac{2}{7}=p$ so $\frac{7p}{24}=\frac{1}{12}=0.83333333333$ which is the answer