SOLUTION: Urn A contains six white balls and four black balls. Urn B contains five white balls and three black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is then

Question 1130319: Urn A contains six white balls and four black balls. Urn B contains five white balls and three black balls. A ball is drawn from Urn A and then transferred to Urn B. A ball is then drawn from Urn B. What is the probability that the transferred ball was white given that the second ball drawn was white? (Round your answer to three decimal places.)
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

I will assume, since you asked this question, that you are not an expert in probability. In that case, I would strongly recommend that you calculate the probabilities of all possible outcomes:

(1) the transferred ball is white, and the selected ball is white
(2) the transferred ball is white, and the selected ball is black
(3) the transferred ball is black, and the selected ball is white
(4) the transferred ball is black, and the selected ball is black

Doing that gives you good practice in performing the required calculations. It also helps you see if you are doing the calculations correctly, because you know that the sum of the probabilities for all four cases must be 1.

And another suggestion.... When doing the calculations for each case, use fractions instead of decimals; and do NOT simplify fractions until you are ready to answer the question that was asked.

Let's go....

(1) the transferred ball is white, and the selected ball is white

6 of the 10 balls in urn A are white; the probability of choosing a white one is 6/10.
With a white transferred to urn B, 6 of the 9 balls now in urn B are white; the probability of selecting a white one is 6/9.
The probability for this first case is (6/10)(6/9) = 36/90.

(2) the transferred ball is white, and the selected ball is black

6 of the 10 balls in urn A are white; the probability of choosing a white one is 6/10.
With a white transferred to urn B, 3 of the 9 balls now in urn B are black; the probability of selecting a black one is 3/9.
The probability for this second case is (6/10)(3/9) = 18/90.

Notes to help you when solving other similar problems:
(a) The first probability fraction for cases 1 and 2 is the same; in both cases a white is transferred from urn A to urn B
(b) The SUM of the second probability fractions for cases 1 and 2 is 1, because the ball selected from urn B has to be either white or black.

(3) the transferred ball is black, and the selected ball is white

4 of the 10 balls in urn A are black; the probability of choosing a black one is 4/10.
With a black transferred to urn B, 5 of the 9 balls now in urn B are white; the probability of selecting a white one is 5/9.
The probability for this third case is (4/10)(5/9) = 20/90.

(4) the transferred ball is black, and the selected ball is black

4 of the 10 balls in urn A are black; the probability of choosing a black one is 4/10.
With a black transferred to urn B, 4 of the 9 balls now in urn B are black; the probability of selecting a black one is 4/9.
The probability for this fourth case is (4/10)(4/9) = 16/90.

Notes to help you when solving other similar problems:
(a) The first probability fraction for cases 3 and 4 is the same; in both cases a black is transferred from urn A to urn B
(b) The SUM of the second probability fractions for cases 3 and 4 is 1, because the ball selected from urn B has to be either white or black.
(c) The SUM of the first probability fraction for cases 1 and 2 and the first probability fraction for cases 3 and 4 is 1, because the ball transferred from urn A to urn B has to be either white or black.

At this point, check to see that the sum of the probabilities is 1:

36/90 + 18/90 + 20/90 + 16/90 = 90/90 = 1

Now to answer the question that was asked, using these probabilities:

What is the probability that the transferred ball was white given that the second ball drawn was white?

The phrase "given that the second ball drawn was white" means that the "sample space" is only the outcomes in which the ball selected from urn B was white -- cases 1 and 3. That means the denominator of the probability fraction is the sum of the probabilities of those two cases: 36/90+20/90= 56/90.

And of those outcomes in which the ball drawn from urn B was white, the favorable outcomes are the ones in which the ball transferred from urn A to urn B was white -- case 1. So the numerator of the probability fraction is the probability of just case 1: 36/90.

So the probability we are asked to find is

(36/90)/(56/90) = 36/56 = 9/14

Note that the work was made easier by NOT simplifying fractions until the final answer.

And it is unfortunate that the problem asks you to round the answer to 3 decimal places, because that answer will not be exact, whereas the answer "9/14" IS exact.

9/14 = 0.643 to 3 decimal places

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