Question
Hey everybody I am having trouble with thisproblem:
An IRS agent randomly selects 15 income tax returns for audit. (Sheaudits 15 returns each day.) In the past, the agent has found that20% of her audited returns contain errors.
a) Find the probability that for oneparticular day, there are at least two returns with errors.
b) If the agent works on a weekend and audits 30 returns, find theprobability that there will be exactly 4 returns witherrors.
Thank You, Please show all work, Will give aLifesaver. Thanks
Explanation / Answer
This is a binomial distribution. Each of the audits hasa 0.2 chance of having an error. . The probability of there being no errors is 0.8^15. . The probability of there being exactly one error, and itsbeing the first audit, is 0.2*0.8^14. The probability ofthere being exactly one error, and its being the second audit, is0.8*0.2*0.8^13. And so on, so that the probability of therebeing exactly one error is 15*0.2*0.8^14 . This should start to warn you that this isa combinatorics problem. The probability of there beingexactly two errors, occurring in very specific places, is0.2^2*0.8^13. How many different places are there to putthose 2 errors? From the group of 15 possible places to putthe errors, we want to choose 2 of them, so how many combinationsare there of 15-choose-2? The formula is 15!/(13!2!) . If you don't remember your combinatorics, now is agood time to review.... . The general formula for the probability of having exactly kerrors out of n audits, when the probability of an error is p,is . [n!/(n-k)!k!)] * p^k * (1-p)^(n-k) . We call this the BINOMIAL DISTRIBUTION. . Once you have the formula, the rest is plugging andchugging. . HINT: to find the probability that there are AT LEAST tworeturns with errors could be a real pain in the neck, since you'dhave to find the probability that k=2, k=3, k=4, etc, all the wayup to k=15, and then add them all together. A much faster wayis to find the probability that LESS THAN two returns have errors,which just requires plugging in k=0 and k=1. (You don't evenneed the factorials for those! The number of combinationswith k=0 is one, and with k=1 is 15, as discussed above.) Then subtract this probability from 1. Muchsimpler.