Question

There are 4 urns. The first urn contains 1 white & 1 black ball, the second urn contains 2 white & 3 black balls, the third urn contains 3 white & 5 black balls & the fourth urn contains 4 while & 7 black balls. The selection of each urn is not equally likely. The probability of selecting $i^{th}$ urn is $\frac{i^2+1}{34}(i=1,2,3,4)$. If we randomly select one of the urns & draw a ball, then the probability of ball being white is $\frac{p}{q}$ then sum of digits of p is. (Where p & q are co-prime natural numbers) :

Answer 1

Given:

Urn 1: $U_1$ has 1 white and 1 black balls, total balls = 2. Hence probability of white balls is , $P\left(\frac{W}{U_1}\right)=\frac{1}{2}$

Urn 2: $U_2$ has 2 white and 3 black balls., total balls = 5. Hence probability of white balls is , $P\left(\frac{W}{U_2}\right)=\frac{2}{5}$

Urn 3: $U_3$ has 3 white and 5 black balls., total balls = 8. Hence probability of white balls is , $P\left(\frac{W}{U_3}\right)=\frac{3}{8}$

Urn 4: $U_4$ has 4 white and 7 black balls., total balls = 11. Hence probability of white balls is , $P\left(\frac{W}{U_4}\right)=\frac{4}{11}$

And $P\left(i^{th}urn\right)=\frac{i+1}{34}(i=1,2,3,4)$

To find:

Total P(white ball)

$P\left(W\right)=P\left(U_1\right)\times P\left(\frac{W}{U_1}\right)+P\left(U_2\right)\times P\left(\frac{W}{U_2}\right)+P\left(U_3\right)\times P\left(\frac{W}{U_3}\right)+P\left(U_4\right)\times P\left(\frac{W}{U_4}\right)$

$P\left(W\right)=\frac{2}{34}\times \frac{1}{2}+\frac{5}{34}\times \frac{2}{5}+\frac{10}{34}\times \frac{3}{8}+\frac{17}{34}\times \frac{4}{11}⟹P\left(W\right)=\frac{1}{34}+\frac{2}{34}+\frac{15}{4\times 34}+\frac{68}{11\times 34}⟹P\left(W\right)=\frac{44+88+165+272}{4\times 11\times 34}=\frac{569}{1496}$

Given the probability of ball being white is $\frac{p}{q}$

Hence $p=569$

Therefore sum of digits of p is $5+6+9=20$