# Lewis Carroll Probability Problem

• May 3rd 2011, 04:23 AM
SteveSmith
Lewis Carroll Probability Problem
This problem was included in a sermon on Sunday by a Canon at St Georges Chapel (the private chapel inside Windsor Castle, UK). It apparently originates from Lewis Carroll, author of Alice in Wonderland. The Canon gave Carroll's answer to the problem, but I got a different answer, so I wanted to get some expert opinions. The problem is :

There are two balls in a bag. Ball A is known to be white, ball B is unknown, but is equally likely to be either black or white. A ball is drawn at random from the bag, and is white. What is the probability that the ball remaining in the bag is white?

The canon's (and Lewis Carroll's) answer was two thirds, based on the following logic :

There are four possible outcomes here :

1) Ball A has been withdrawn, ball B (white) remains
2) Ball A has been withdrawn, ball B (black) remains
3) Ball B (white) has been withdrawn, ball A remains
4) Ball B (black) has beenwithdrawn, ball A remains

Since the ball that has been withdrawn is white, option 4 is ruled out. There are therefore three remaining options, of which two (1 and 3) have a white ball remaining in the bag, so the probability is two thirds.

I'm not convinced this is right, since I don't think the three remaining options are equally likely. My answer was 75%, based on the following :

The white ball that has been withdrawn is equally likely to be either A or B, so 50% chance of each. There are two ways for a white ball to remain in the bag.

If the withdrawn ball is B, then the remaining ball must be A, which is white, so the probability of white in this case is 50% x 100% = 50%.

If the withdrawn ball is A, then there is a 50% chance that the remaining ball B is white, so the probability in this case is 50% x 50% = 25%.

The total probability of a white ball remaining is therefore 50% + 25% = 75%

Which of us (if either!) is correct?

Thanks,

Steve.
• May 3rd 2011, 04:46 AM
SpringFan25
The white ball that is drawn first is not equally likely to be B or A. Given that the first ball is white, there is a 2/3 chance that it is A. (draw a tree diagram to check this)

My thinking:
P(second ball is white) = P(both were white|at least one is white) = P(both were white)

You are given this probability as 50%.

But
im not sure, which is actually pretty alarming but never mind.
• May 4th 2011, 10:17 AM
SpringFan25
oops. i was wrong, it is 2/3.