Probability help!! please!!
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Probability help!! please!!

[From: ] [author: ] [Date: 11-05-26] [Hit: ]
21)A) prob at most 3 customers under $50k audited = p(0 audits) + p(1 audit) + p(2 audits) + p(3 audits) p(0) = (1-0.12)^45 = 0.88^45 = 0.0032 p(1) = 0.12 * 0.88^44 * 45 = 0.......
According to a well-known accounting firm, the chances of your tax return being audited by Revenue Canada are about 12 in 100 if your income is less than $50,000. The chances increase to 21 in 100 if your income is more than $50,000. You are presently employed by an accounting firm and have prepared tax returns for 45 clients with income under $50,000 and for 35 clients with income over $50,000.
a. What is the probability that at most 3 of the clients with incomes under $50,000 will be audited?
b) What is the probability that at least 5 of your clients that have incomes over $50,000 will be audited?

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p(audit):
(income < $50,000 / 45 clients) = 12% ( or 0.12)
(income > $50,000 / 35 clients) = 21% ( or 0.21)

A) prob at most 3 customers under $50k audited
= p(0 audits) + p(1 audit) + p(2 audits) + p(3 audits)
p(0) = (1-0.12)^45 = 0.88^45
= 0.0032
p(1) = 0.12 * 0.88^44 * 45
= 0.0195
p(2) = 0.12^2 * 0.88^43 * 45C2 (45C2 = 45!/(43!2!) = (45*44)/(2*1)
= 0.0584
p(3) = 0.12^3 * 0.88^42 * 45C3 (45C3 = (45*44*43)/(3*2*1) )
= 0.1005

-> P(max 3 audits) = 0.1816


B) prob at least 5 customers over $50k audited
= 1 - p(max 4 audits)
using same principes as above:
p(0) = 0.79^35
= 0.0002
p(1) = 0.21 * 0.79^34 * 35
= 0.00005
p(2) = 0.21^2 * 0.79^33 * 35C2
= 0.0110
p(3) = 0.21^3 * 0.79^32 * 35C3
= 0.0254
p(4) = 0.21^4 * 0.79^31 * 35C4
= 0.0683

--> p(max 4 audits) = 0.1049

--> p(at least 5 audits ) = 1 - 0.1049 = 0.8951

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I'm not sure if someone will solve this question with the information you are letting us know. You could explain a little more so that we can help with more confidence. Just a thought! :-D
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keywords: please,help,Probability,Probability help!! please!!
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