Question

Urn I contains two white chips and one red chip; urn II has one white chip and two red chips. One chip is drawn at random from urn I and transferred to urn II. Then one chip is drawn from urn II. Suppose that a red chip is selected from urn II. What is the probability that the chip transferred was white?

Answer 1

The probability is 0.5714

Explanation:

Let's call W the event that the chip transferred is white, W' the event that the the chip transferred is red and R the event that a red chip is selected from urn II.

So, the probability that the chip transferred was white given that a red chip is selected from urn II is:

P(W/R) = P(W∩R)/P(R)

Where P(R) = P(W∩R) + P(W'∩R)

Therefore, the probability that the chip transferred is white is P(W) = 2/3, because on urn I, there are 3 chips and 2 of them are white.

If the chip transferred is white, the probability of select a red chip from urn II is P(R/W) = 2/4, because there are now 4 chips on urn II and 2 of them are red.

Finally, P(W∩R) = P(W)*P(R/W) = 2/3*2/4 = 1/3 = 0.3333

At the same way, The probability that the chip transferred is red is P(W') = 1/3 and the probability of select a red chip from urn II given that the chip transferred is red is P(R/W') = 3/4.

Finally, P(W'∩R) = P(W')*P(R/W') = 1/3*3/4 = 1/4 = 0.25

So, P(R) and P(W/R) are calculated as:

P(R) = 0.3333 + 0.25 = 0.5833

P(W/R) = 0.3333/0.5833 = 0.5714