Paychecks and Envelopes
Date: 10/07/97 at 12:01:10 From: Babette Martin Subject: Probability Five paychecks and envelopes are addressed to five different people. If the paychecks are randomly inserted into the envelopes, what is the probability that (a) exactly one will be inserted in the correct envelope, and (b) at least one will be inserted in the correct envelope? We were able to figure the problem out when we only had four checks and envelopes because we could list all the possible combinations. It is almost impossible for us to list the 120 different combinations for the 5 paychecks. Help - we need a better way to figure this problem out! We are anxiously waiting for your suggestion. Thanks!
Date: 10/07/97 at 16:25:56 From: Doctor Anthony Subject: Re: Probability If we let A be the event that paycheck A is in correct envelope and similarly B is event that paycheck B is in correct envelope, then P(A) = 1/5 P(A and B) = 1/5 x 1/4 Now use the inclusion, exclusion formula 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 x (1/5) - 5C2 x (1/5)(1/4) + 5C3 x (1/5)(1/4)(1/3) - 5C4 x (1/5)(1/4)(1/3)(1/2) + (1/5)(1/4)(1/3)(1/2)(1/1) 5x4 1 5x4x3 1 5x4x3x2 1 1 = 1 - --- x --- + ----- x ----- - ------- x ------- + -------- 1x2 5x4 1x2x3 5x4x3 1x2x3x4 5x4x3x2 5x4x3x2x1 = 1 - 1/2! + 1/3! - 1/4! + 1/5! This is a probability that at least one paycheck is correctly placed. The chance that none is correctly placed is 1 - (above result) = 1 - 1 + 1/2! - 1/3! + 1/4! - 1/5! = 1/2! - 1/3! + 1/4! - 1/5! To find the probability that exactly one is correctly placed, we need to find the total number of ways with 1 correct and 4 incorrect. We can use the result of last section to see the number of ways where 4 checks are all incorrect. If we had 4 checks and 4 envelopes, the probability that all are incorrect would be 1/2! - 1/3! + 1/4! Therefore the number of arrangements with all incorrect is given by 4!(1/2! - 1/3! + 1/4!) = 12 - 4 + 1 = 9 Now imagine the checks laid out in a row, and the envelopes in a matching row. If the first check is the only correct one opposite its envelope, then there are 9 arrangements for the other 4 all putting them in the incorrect envelope. Similarly, if the second one is the only correct one, then this can be associated with 9 arrangements for the other 4, which will all be incorrect. Similarly for the third, fourth, and fifth checks, they can each be associated with 9 arrangements, making the other 4 all incorrect. So the total number of arrangements in which just one is correctly placed is: 5 x 9 = 45 The total number of possible arrangements is of course 5! = 120, so the probability that exactly one is correct is 45/120 = 3/8 We have already seen that the probability of at least one being correct is: 1 - 1/2! + 1/3! - 1/4! + 1/5! = 19/30 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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