Not the First Event and Not the Second — and Not IndependentDate: 07/20/2012 at 22:29:05 From: Brian Subject: probability word problem At a card party, two door prizes are to be awarded by drawing two names at random from a box. The box contains the names of 40 people, including that of Joe Ramirez. What is the probability that Joe's name is not drawn for either prize? I reason that there is a 1/40 chance for Joe's name to be selected on the first draw, and a 1/39 chance on the second draw. The probability of Joe getting a prize on one of the draws is the sum of these, or 79/1560. The probability of him not getting a prize is therefore 1 - 79/1560, or 1481/1560. Is this correct? The answer key gives 1482/1560, or 19/20. Date: 07/21/2012 at 12:59:28 From: Doctor Peterson Subject: Re: probability word problem Hi, Brian. The trouble is that the addition rule you are following doesn't have the special case that the multiplication rule has, in which you can treat the event "name is drawn on second draw" as a conditional probability (that is, ASSUME that the name was not drawn the first time). To find the actual probability that his name is drawn second, you would have to do this: P(drawn second) = P(not drawn first AND drawn second) = P(not drawn first) * P(drawn second | not first) = 39/40 * 1/39 = 1/40 That is, if you don't know what happened on the first draw, it makes no difference whether you ask about the first or the second draw! That makes sense when you think about it, but I did it this way to convince you (and myself), because it's not entirely obvious that it's not a fallacy. So the probability you calculated is really P(drawn first OR drawn second) = P(first) + P(second) - P(both) = 1/40 + 1/40 - 0 = 1/20 By complementarity, P(neither first nor second) = 1 - P(first or second) = 1 - 1/20 = 19/20 The usual way is to treat the problem as an "and," using the rule for non-independent events: P(neither first nor second) = P(not first AND not second) = P(not first) * P(not second|not first) = 39/40 * 38/39 = 38/40 = 19/20 - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 07/21/2012 at 19:13:40 From: Brian Subject: Thank you (probability word problem) You identify exactly where I went wrong in your comment that "the addition rule you are following doesn't have the special case that the multiplication rule has, in which you can treat the event 'name is drawn on second draw' as a conditional probability." Framing the problem this way made the correct solution crystal clear: P(drawn second) = P(not drawn first AND drawn second) Thanks, Dr. Peterson! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/