Question

Box 1 contains 2 yellow and 4 green balls, whereas Box 2 contains 1 yellow and 1 green ball. A ball is randomly chosen from Box 1 and then transferred to Box 2, and a ball is then randomly selected from Box 2. a. What is the probability that the ball selected from Box 2 is yellow? b. What is the conditional probability that the transferred ball was yellow, given that a yellow ball is selected from Box 2?

Answer 1

Answer:

Step-by-step explanation:

Given

Box 1 contains 2 yellow and 4 green balls

box 2 contains 1 yellow and 1 green ball

case 1 when yellow ball is chosen and transferred to second box thus Probability associated with it

$P_1=\frac{2}{6}\times \frac{2}{3}=\frac{4}{18}$

case 2 when Green ball is chosen and transferred to second box thus Probability associated with it

$P_2=\frac{4}{6}\times \frac{1}{3}=\frac{4}{18}$

Thus Probability that ball selected form Box 2 is yellow

$P=P_1+P_2=\frac{4}{18}+\frac{4}{18}=\frac{8}{18}$

(b) P(transferred ball is yellow|Yellow ball is selected)

$P(Yellow\ ball\ is\ selected)=\frac{8}{18}$

$P(transferred\ ball\ is\ yellow|Yellow\ ball\ is\ selected)=\frac{Probability\ of\ transferring\ yellow\ ball}{Probability\ of\ selecting\ Yellow\ ball}$

$P=\frac{\frac{4}{18}}{\frac{8}{18}}=0.5$