Choosing Chocolate Bars
Date: 12/04/2001 at 10:49:13 From: Tracie Berlin Subject: Probability If I have 3 Mr. Goodbars, 13 Hershey milk chocolate bars, 7 Special Dark bars, and 3 Krackel bars, what is the probability that I will get at least 1 Krackel if I pull two out of the bag at the same time (together, not looking at one first and then looking at the other)? I know the probability of pulling a Krackel out if I pick one piece of candy is 3/26, but I don't know if I can even figure out how you'd do it if you had two. You know the chances are greater, but I'm not sure if you can just take 3/26 + 3/26, because there would only be 25 after the first has been taken out. You have no way of knowing what that first bar is because you are looking at them both at the same time. I'm stumped. Help!
Date: 12/04/2001 at 12:21:12 From: Doctor Greenie Subject: Re: Probability Hello, Tracie - This is a nice problem to show two alternative ways to attack the same problem. These two alternative approaches are based on the following two different ways of rephrasing "... at least one Krackel..." (1) "at least one Krackel" = "either one Krackel or two Krackels" (2) "at least one Krackel" = "not zero Krackels" With the first interpretation, the "... either ... or ..." means you find the answer by adding the probabilities of "one Krackel" and "two Krackels." With the second interpretation, you find the probability of "zero Krackels", and the answer to the problem is "everything else" - i.e., 1 minus the probability of zero Krackels. Let's solve the problem both ways. You are right about not being able to use 3/26 twice, because once one bar is selected there are no longer 26 bars to choose from for the second bar. (Note that even though you say you choose two at the same time, mathematically we still consider the two choices as occurring one after the other....) So, for example, if we consider the case of getting two Krackels, the probability of getting a Krackel with the first choice is 3/26, and the probability of getting another Krackel with the second choice is 2/25. First interpretation. In this interpretation, "at least one Krackel" is thought of as "either one or two Krackels"; thinking of the two selections as occurring one after the other, this is then equivalent to "either Krackel and then another Krackel, or Krackel and then something else, or something else and then Krackel." 3 2 P(Krackel, Krackel) = -- * -- 26 25 3 23 P(Krackel, other) = -- * -- 26 25 23 3 P(other, Krackel) = -- * -- 26 25 So we then have 3 2 3 23 23 3 P(at least one Krackel) = -- * -- + -- * -- + -- * -- 26 25 26 25 26 25 6+69+69 144 = ------- = --- 650 650 Second interpretation. In this interpretation, "at least one Krackel" is thought of as "not zero Krackels". 23 22 506 P(other, other) = -- * -- = --- 26 25 650 So we then have P(at least one Krackel) = 1 - P(other, other) 506 144 = 1 - --- = --- 650 650 I hope this helps. Write back if you have further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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