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Q: probability for 4 balls thrown into 4 boxes ( Answered ,   0 Comments )
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 Subject: probability for 4 balls thrown into 4 boxes Category: Science > Math Asked by: emmag999-ga List Price: \$25.00 Posted: 29 Apr 2006 12:05 PDT Expires: 29 May 2006 12:05 PDT Question ID: 723962
 ```Hi Have been going through past exam papers in statistics and have got completely stuck on this one. Four balls are thrown at random into 4 boxes, with each ball having an equal chance of going into every box. Work out the probability that exactly m boxes are empty, where m = 0,1,2,3. I managed m=0 and m=3 fine, and I have managed to work the other 2 out the long way round by drawing it all out, but the answer given in the exam report states probability that 1 box is empty is 12 x [4!/(2!1!1!0!)]x 1/4squared x 1/4 x 1/4 = 36/64 and I have no idea how they got this. Please help?```
 ```Dear Emma, Although in principle you can solve problems of this kind by enumerating all possibilities on paper, in practice it is unlikely that an exam will permit you enough time to do so. It is better to compute the probability by an analysis that minimizes counting. Among the various analytical methods one may employ to arrive at the answer, the simplest begins by computing the total number of ways there are to distribute four distinct balls among four distinct boxes. Observe that we can place the first ball into one of four boxes, the second ball again into one of four, the third into one of four, and the fourth into one of four. This makes for a total of 4*4*4*4 = 256 possibilities. Next, we compute the number of cases in which exactly one box is empty. Dividing this number by 256 will give us the desired probability. There are four factors to consider. 1. If exactly one box is empty, there are four possibilities as to which one it is. factor: 4 2. Of the remaining three boxes, exactly one must contain two balls. This gives a further three possibilities. factor: 3 3. How many possibilities are there for the two balls that go into one box? This number is written C(4, 2), pronounced "4 choose 2", and is conventionally computed as 4! 4*3*2*1 4*3 ---- = ------- = --- = 6 . 2!2! 2*1*2*1 2*1 This just says that we pick one of four balls and then one of three, for a total of 4*3 = 12 permutations. But half of those permutations are mirror images -- the same two balls in reverse order -- so we divide by two, giving us 12/2 = 6 combinations. factor: 6 4. Finally, we consider the two remaining balls in their two individual boxes. There are exactly two ways to order these two balls: left to right, and right to left. factor: 2 We multiply the four factors to find that there are 4*3*6*2 = 12*12 = 144 cases in which exactly one box is empty. Dividing by the total number of cases, we obtain a probability of 144 72 36 18 9 --- = --- = -- = -- = -- . 256 128 64 32 16 So 36/64 is one way of expressing the probability, but the simplest fraction is actually 9/16. Regards, leapinglizard```