Date: 05/05/98 at 05:32:29 From: Veronica West Subject: Picking the same number Dear Dr. Math: A jar contains 19 different marbles. I draw 13 marbles, one at a time, but I put each marble back in the jar before taking the next. What is the possibility that, out of the 13 times, I will take any marble twice? (Not a specific marble, but just one of them, and I do not have to take them just after each other). What is the possibility of taking the same marble 5 times out of the 13 times (in the same way as above)? I hope you can help me with these problems!
Date: 05/05/98 at 12:09:09 From: Doctor Sam Subject: Re: Picking the same number Veronica, A confusing question. Let's see if I can help. Since the marbles are all different, let's think of them as being numbered 1 through 19. Since you return the marble to the jar each time, the probability of picking any particiular marble is always 1/19. It may help to think about drawing the thirteen marbles in a different way. If we pick a marble and record its number, then our thirteen choices will produce a sequence of thirteen numbers, each between 1 and 19. Instead of thinking of picking marbles, then, we can think about making sequences of thirteen numbers. How many sequences can we make that will have a repeated number? Well, most of the sequences will have one or more repeated numbers. This is the kind of question that is best answered in reverse: "How many sequences will have no number repeated?" This is just P(19, 13), which equals the number of permutations of nineteen numbers taken thirteen at a time. You can also get this value by using the Multiplication Principle. There are 19 possible choices for the first number but only 18 choices for the second number, only 17 choices for the third number, and so on. Altogether, there are: 19 * 18 * 17 * 16 * 15 * 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 such sequences, that is, sequences with NO repeated number. (This can be written more simply as 19!/4!) All other sequences of 13 numbers will have some repetition. So how many sequences are there altogether? There are 19 possible choices for the first number AND 18 choices for the second number AND 19 choices for the third number, ..., for a total of 19^13 choices. Therefore, there are 19^13 - 19!/4! sequences that have some repetition. What is the probability of picking one of these? Well, we said that the probability of picking any particular number (marble) is 1/19. So the probability of picking thirteen numbers to match a particular sequence is: (1/19)*(1/19)*(1/19)*...*(1/19) = 1/19^13. The probability of picking one of these sequences is, therefore: 19^13 - 19!/4! -------------- 19^13 Now this is not a direct answer to either of your questions, but if you understood the idea behind what I did then the rest should be a lot easier. QUESTION 1: Exactly 1 marble (number) is repeated twice. How many sequences will have exactly one number repeated twice? We can form the sequence by: 1) choosing the number that will be duplicated 2) choosing where in the sequence the repetitions will occur 3) choosing the other 11 numbers There are 19 choices for (1) since any of the marbles may be the target. There are C(13, 2) = the number of combinations of 13 things taken 2 at a time = 13 * 12 / 2 = 78. We need this value because the repetition could occur right away with the first two marbles we choose, or it might be a repetition of the seventh marble chosen, or it might be some other variation. In each case we can think of this as a sequence of 13 objects, and we want two positions to be the same. That is a combinations question. Since only this number is to be repeated, all other numbers must be different. We already chose the repeated number, so that leaves 18 numbers to choose from for the remaining 11 positions in the sequence. Since we want them to be different, this is P(18,11) = 18!/11! The number of such sequences is 19 * 78 * 18! / 11! The probability of picking one of these sequences is: (19 * 78 * 18! / 11!)/19^13 because, as we said, the probability of picking ANY sequence of 13 marbles is 1/19^13. QUESTION 2: Now we want one marble to be repeated 5 times. The reasoning is exactly the same as in (1) except that we now want five repetitions. That means the sequence will include five numbers that are the same. Here are the steps in creating such a sequence: 1) choose the number that will be duplicated 2) choose where in the sequence the 5 repetitions will occur 3) choose the other 8 numbers The number of such sequences is: 19 13! 18! 19! (19)*C(13, 5)*P(18, 5) = ----------- = ----- 5! 8! 13! 5! 8! And the probability of picking one of these is: 19! ---------- 8! 19^13 I hope that helps! -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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