Question

Let $u_1$ and $u_2$ be two urns such that $u_1$ contains 3 white, 2 red balls and $u_2$ contains only 1 white ball. A fair coin is tossed. If head appears, then 1 ball is drawn at random from urn $u_1$ and put into $u_2$. However, if tail appears, then 2 balls are drawn at random from $u_1$ and put into $u_2$. Now, 1 ball is drawn at random from $u_2$. Then, probability of the drawn ball from $u_2$ being white is

• A. $\frac{13}{30}$
• B. $\frac{23}{30}$
• C. $\frac{19}{30}$
• D. $\frac{11}{30}$

Answer 1

Answer: (B)

$\frac{23}{30}$

Case 1: head, white from $U_1$, white from $U_2=\left(1/2\right)\left(3/5\right)\left(2/2\right)=3/10$

Case 2: head, red from $U_1$, white from $U_2=\left(1/2\right)\left(2/5\right)\left(1/2\right)=1/10$

Case 3: tail, 2 white from $U_1$, white from $U_2=\left(1/2\right){(}^3{C}_2{/}^5{C}_2\left)\right(3/3)=3/20$

Case 4: tail, white and red from $U_1$, white from$U_2=\left(1/2\right){(}^3{C}_1{\ast }^2{C}_1{/}^5{C}_2\left)\right(2/3)=1/5$

case 5: tail, 2 red fom $U_2$, white from $U1=\left(1/2\right){(}^2{C}_2{/}^5{C}_2\left)\right(1/3)=1/60$

Required Probability= sum of probabilities of above cases$=\left(3/10\right)+\left(1/10\right)+\left(3/20\right)+\left(1/5\right)+\left(1/60\right)=46/60=23/30$