Question

Box I contains three blue and two red marbles, Box II contains three blue and three red marbles. Suppose one marble is drawn from Box I and placed in Box II and then one marble is drawn from Box II. Find the conditional probability that the marble drawn from Box I is blue given that the marble drawn from Box II is red.

Answer 1

Let P(B1) and P(R1) be the probabilities to draw blue and red balls resepectively from Box I respectively.

P(B1) = 3/5 and P(R1) = 2/5

Similarly, let P(B2) and P(R2) be the probabilities to draw blue and red balls resepectively from Box II respectively.

The conditional probability that the marble drawn from Box I is blue given that the marble drawn from Box II is red is, P(B1 | R2)

By Bayes theorem,

P(B1 | R2) = P(R2 | B1) * P(B1) / P(R2)

Given, blue ball is drawn from Box I, then total number of blue balls in Box II will become 4.

Then,   P(R2 | B1) = 3 / 7

Given, red ball is drawn from Box I, then total number of red balls in Box II will become 4.

Then,   P(R2 | R1) = 4 / 7

By law of total probability,

P(R2) = P(R2 | B1) P(B1) + P(R2 | R1) P(R1)

= (3/7) * (3/5) + (4/7) * (2/5)

= (9/35) + (8/35)

= 17/35

Now, P(B1 | R2) = P(R2 | B1) * P(B1) / P(R2)

= (3/7) * (3/5) / (17/35)

= 9/17

= 0.5294118