Problem: IRS will audit 2 million taxpayers. Plus 1 million taxpayers will be audited by mail. Assume 100 million total taxpayers. Find the probability that in a random sample of 50 taxpayers there will be one audit and one mail-audit. My approach: For the audit: Used Poisson distribution with mu = 1 and x = 1. P(1) = 1^1 * e^(-1) / 1! = 0.368. For the mail-audit, I assumed the audit and mail-audit are independent, so I can use Poisson again with mu = 0.5 and x = 1. I get 0.303. For both events, multiply, yielding 0.112. What if I assume that the IRS will exclude auditees from mail-a

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Problem: IRS will audit 2 million taxpayers. Plus 1 million taxpayers will be audited by mail. Assume 100 million total taxpayers. Find the probability that in a random sample of 50 taxpayers there will be one audit and one mail-audit. My approach: For the audit: Used Poisson distribution with mu = 1 and x = 1. P(1) = 1^1 * e^(-1) / 1! = 0.368. For the mail-audit, I assumed the audit and mail-audit are independent, so I can use Poisson again with mu = 0.5 and x = 1. I get 0.303. For both events, multiply, yielding 0.112. What if I assume that the IRS will exclude auditees from mail-a

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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hmm , does the problem say to use Poisson
Not explicitly, but based on the section where the problem was presented and the rules given in the text, it seems that Poisson was an acceptable approximation to the Binomial. n = 50. Although p = 0.01 is a little close to zero.

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