Probability of Matching Envelopes and LettersDate: 03/03/98 at 04:20:34 From: Raveen Dey Subject: Probability- permuatations Six letters were to be placed in envelopes for posting. Unfortunately, the letters were dropped before being put in their envelopes, and they were placed in at random. a) What is the probability that none is in the correct envelope? b) What is the probability that exactly one is in the correct envelope? Answers a)53/144 b)11/30 Date: 03/03/98 at 08:14:40 From: Doctor Anthony Subject: Re: Probability- permutations I answered a nearly identical question earlier today, so I will copy that answer here. I was considering 5 letters and 5 envelopes. If we let A be the event that letter A is in ist correct envelope and similarly B is the event that letter B is in its correct envelope, then P(A) = 1/5 and P(A and B) = 1/5 * 1/4 Now use the inclusion-exclusion principle to get probability that A or B or C .... or E are correctly placed. P(A or B or C .... or E) = P(A) + P(B) + P(C) + P(D) + P(E) - P(A and B) - P(B and C) -.... + P(A and B and C) + P(B and C and D) + .... - P(A and B and C and D) - P(...) -...... + P(A and B and C and D and E) = 5*(1/5) - 5C2*(1/5)(1/4) + 5C3*(1/5)(1/4)(1/3) - 5C4*(1/5)(1/4)(1/3)(1/2) + (1/5)(1/4)(1/3)(1/2)(1/1) 5*4 1 5*4*3 1 5*4*3*2 1 1 = 1 - --- * --- + ----- * ----- - --------- * ------- + --------- 1*2 5*4 1*2*3 5*4*3 1*2*3*4 5*4*3*2 5*4*3*2*1 = 1 - 1/2! + 1/3! - 1/4! + 1/5! This is the probability that at least one letter is correctly placed. The chance that none is correctly placed is 1 - (above result), or = 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! = 1/2! - 1/3! + 1/4! - 1/5! With n letters and envelopes, the probability that none is correctly placed is: = 1/2! - 1/3! + 1/4! - 1/5! + ....... + (-1)^n 1/n! Note that as n becomes very large, this probability tends to the value e^(-1), since: e^(-1) = 1 - 1 + 1/2! - 1/3! + 1/4! - ........ to infinity = 0.367879... With six letters and envelopes, the probability that none is correctly placed is: = 1/2! - 1/3! + 1/4! - 1/5! + 1/6! = 1/2 - 1/6 + 1/24 - 1/120 + 1/720 = 53/144. To find the probability that exactly one is correctly placed, we need to find the total number of ways with 1 correct and 5 incorrect. We can use the result of the last problem to see the number of ways where 5 letters are all incorrect. If we had 5 letters and 5 envelopes, the probability that all are incorrect is 1/2! - 1/3! + 1/4! - 1/5! Therefore the number of arrangements with all incorrect is given by 5!(1/2! - 1/3! + 1/4! - 1/5!) = 44 Now imagine the letters laid out in a row, and the envelopes in a matching row. If the first letter is the only correct one opposite its envelope, then there are 44 arrangements for the other 5 all putting them in the incorrect envelope. Similarly, if the second one is the only correct one, then this can be associated with 44 arrangements for the other 5, which will all be incorrect. Similarly, for third, fourth, fifth and sixth letters, they can each be associated with 44 arrangements, making the other 5 all incorrect. So the total number of arrangements in which just one is correctly placed is: 6 x 44 = 264 The total number of possible arrangements is of course 6! = 720, so the probability that exactly one is correct is 264/720 = 11/30 -Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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