Probability Question from a Math TestDate: 3/8/96 at 21:9:23 From: Anonymous Subject: Question Dear Dr. Math, I have a question about a math problem that came up on a math test I took recently. There are 15 homerooms in the school. There are 20 students in each homeroom. If the principal selects 5 of the homerooms for a pizza party, what is the probability of Mr. Smith's homeroom being selected? a. 1/3 b. 1/5 c. 15/20 d. 1/20 Can you help me? I feel the answer is a. 1/3, but why is it not 1/15 * 14 * 1/13 * 1/12 * 1/11? And is this a way to find the probability of anything, and if so what would the question be? Date: 3/21/96 at 21:4:52 From: Doctor Aaron Subject: Re: Question Your feeling is right, if we pick 5 out of 15 rooms the probability is 1/3. You could be thinking about several things by the multiplication 1/15 * 1/14 * 1/13 * 1/12 * 1/11. One probability question that this would answer is: What is the probability of the principal picking 5 specific rooms in a specific order? The first time (s)he must pick one room out of 15, then out of 14, and so on, so the probability of the principal picking all 5 rooms in order is the product of the probability of picking each room individually, i.e., 1/15 * 1/14 * 1/13 * 1/12 * 1/11. A more common question would ask: what is the chance of the principal picking 5 specific rooms in no particular order. In this case, the probability is that of picking one of the five out of a total of 15, times picking one of the remaining four out of the remaining total of 14, etc. This is (5 * 4 * 3 * 2 * 1)/(15 * 14 * 13 * 12 * 11), an equation of the form: 1/(n!/k!(n-k)!) (where a! = a * a-1 * a-2 * .... * 3 * 2 * 1), which comes up a lot in probability and other areas of math. One name that mathematicians use to refer to the number n!/k!(n- k)! is n choose k, meaning that this represents the number of different combinations of n elements taken k at a time. Here's an example:If we let n = 5 and k = 3, we can think of how many different ways can we take 3 elements from the set {1,2,3,4,5}. We have(1,2,3), (1,2,4), (1,2,5) ... if we write all of them out, we'll have 5!/3!*2! = 10 different combinations. So, had the principal chosen 5 rooms out of 15, (s)he would have had 15 choose 5 choices, from which (s)he chose 1. Then the probability of choosing that 1 set of five rooms would be 1/(15 choose 5) = (5 * 4 * 3 * 2 * 1)/(15 * 14 * 13 * 12 * 11). I hope that this is helpful. -Doctor Aaron, The Math Forum |
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