Cupcakes and Boxes: Conditional Probability
Date: 01/03/2002 at 21:02:23 From: Snowball Subject: Probability Problem a) A mathematics 12 student arrives at a school Math party with three boxes and announces, "These three boxes I am carrying each contain 2 cupcakes - one box has two vanilla, one box has 2 chocolate, and one box has one of each." She the randomly selects a box, opens it and randomly selects a cupcake. The first cupcake is chocolate. If the other cupcake is then pulled from the same box, what is the probability that it will be chocolate? I thought that the probability would be 1/2 because only two boxes have chocolate in them, and in the last box, if the first cupcake is chocolate, the 2nd will have to be vanilla. But my method seems way too easy, as this is a Math 12 Problem designed to be really hard.
Date: 01/04/2002 at 09:49:26 From: Doctor Rick Subject: Re: Probability Problem Hi, Snowball. It is indeed trickier than that. Your answer, as reasonable as it seems, is wrong. There are many probability paradoxes (or rather, surprises) like this, based on the need for precision in defining the random selection process, or on confusion over conditional probabilities. Let's consider all the possible outcomes of selecting a box at random, then a cupcake at random from the box. By this process, each of the following events will have equal probability: 1. First cupcake (vanilla) drawn from first box 2. Second cupcake (vanilla) drawn from first box 3. First cupcake (chocolate) drawn from second box 4. Second cupcake (chocolate) drawn from second box 5. First cupcake (chocolate) drawn from third box 6. Second cupcake (vanilla) drawn from third box Now we select only those events that match the known result: 3. First cupcake (chocolate) drawn from second box 4. Second cupcake (chocolate) drawn from second box 5. First cupcake (chocolate) drawn from third box Each has an equal probability, so each has a 1/3 probability of occurring given that the selected cupcake is chocolate. What will the other cupcake be in each case? 3. Second cupcake in second box is chocolate. 4. First cupcake in second box is chocolate. 5. Second cupcake in third box is vanilla. Thus the probability that the other cupcake is also chocolate is 2/3, not 1/2. Surprise! What is wrong with your answer? You supposed that there was an equal probability that either the second or third box would be chosen. A priori (that is, without the additional knowledge that the cupcake selected is chocolate), this is true; however, with that knowledge, the a posteriori probability is greater that the second box was selected, because there are twice as many ways for the cupcake selected to be chocolate. You can use the formula for conditional probability: P(A|C) = P(A and C)/P(C) In this problem, event A is that the second box was selected, and event C is that a chocolate cupcake was selected. The probabilities are: P(A and C) = 1/3 (= P(A), since both its cupcakes are chocolate) P(C) = 1/2 (3 of the 6 cupcakes are chocolate) Thus, P(A|C) = (1/3)/(1/2) = 2/3 This confirms that the probability that the second box was chosen, given the outcome that a chocolate cupcake was selected from it, is 2/3, not 1/2 as you assumed. You might try calculating the probability that the second cupcake is chocolate given that the first cupcake was chocolate, using conditional probabilities. For fun, you might also take a look at this Dr. Math FAQ: Boy or Girl? http://mathforum.org/dr.math/faq/faq.boy.girl.html - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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