Posted: 1/28/2005 2:32:23 PM EDT
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Question 1: Pollution of the rivers in the United States has been a problem for many years. Consider the following events: A = {The river is polluted} B = {A sample of water tested detects pollution} C = {Fishing permitted} Assume: P(A) = 0.3, P(B l A) = .75 P(B l A') = 0.2 P(C l A n B) = 0.2 P(C l A' n B) = 0.15 P(C l A n B') = 0.8 P(C l A' n B') = 0.9 (a) Find P( A n B n C) 0.045 (b) Find P(B' n C) 0.564 (c) Find P(C) 0.630 (d) Find the probability that the river is polluted, given that fishing is permitted and the sample tested did not detect pollution. 0.1064 The answers in the back of the book are in red. Question 2: A paint store produces and sells latex and semigloss paint. Based on long-range sales, the probability that a customer will buy latex is 0.75. Of those that purchase latex paint, 60% also purchase rollers. But 30% of semigloss buyers purchase rollers. A randomly selected buyer purchases a roller and a can of paint. What is the probability that the paint is latex? |
the easy way to do it, is to find P(b) and P(C), then the rest just sort of falls into place. | means logical 'or', correct?
75% or customers buy latex 60% of 75% buy rollers 25% of customers buy semi-gloss, right? (not the answer, just want to know if that's a valid assumption based on the text, seems to me you don't know if a customer can buy latex and semigloss - it's been a long day). |
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Question 1: P(A)=.3 P(A')=.7 P(B|A) = .75 P(B'|A) = .25 P(B|A') = .2 P(B'|A') = .8 P(B|A) = .75 = P(A n B)/P(A) ---> P(A n B) = (.3)(.75) = .225 P(B|A') = .2 = P(A' n B)/P(A') ---> P(A' n B) = (.7)(.2) = .14 P(B'|A) = .25 = P(A n B')/P(A) ---> P(A n B') = (.3)(.25) = .075 P(B'|A') = .8 = P(A' n B')/P(A') ---> P(A' n B') = (.7)(.8) = .56 P(B) = P(A' n B) + P(A n B) = .225 + .14 = .365 {It's not actually necessary to calculate P(B)} P(C|A n B) = .2 = P(A n B n C)/P(A n B) ---> P(A n B n C) = (.2)P(A n B) = (.2)(.225) = .045 P(C|A' n B) = .15 ---> P(A' n B n C) = (.15)(.14) = .021 P(C|A n B') = .8 ---> P(A n B' n C) = (.8)(.075) = .06 P(C|A' n B') = .9 ---> P(A' n B' n C) = ( .9)(.56) = .504 Part (b): P(B' n C) = P(A' n B' n C) + P(A n B' n C) = .06 + .504 = .564 Part (c): P(C) = P(A n B n C) + P(A' n B n C) + P(A n B' n C) + P(A' n B' n C) = .045 + .021 + .06 + .56 = .63 P(A|B' n C) = P(A n B' n C)/P(B' n C) = .06/.564 = .106383 Question 2: L = {Latex Paint} R = {Roller} P(L) = .75 P(L') = .25 P(R|L) = .6 P(R'|L) = .4 P(R|L') = .3 P(R'|L') = .7 P(L|R) = P(L n R)/P(R) = [P(R|L)P(L)]/P(R) = (.6)(.75)/P(R) = .45/P(R) P(R) = P(R|L)P(L) + P(R|L')P(L') = .45 + (.3)(.25) = .545 P(L|R) = .45/.545 = .825688 |
Never did stats like this, but here's my common sense approach (which may be totally wrong): From the information given (see highlighted stuff in red) twice as many latex purchasers purchase rollers with paint than semigloss paint purchasers. Since the store only sells those two types of paint, the probability is 66.666666% that a purchaser who purchases paint and a roller has purchased latex paint. |
If the probability that a customer buys latex or semigloss is 50% then you would be correct. But the probability of buying latex is 75%. Here another way to think about this problem. The probability of a customer buying a roller and latex is .75*.6 = .45 The probability of a customer buying a roller and semigloss is .25*.3 = .075 Then the probability of a customer buying a roller is .45 + .075 = .525 The probability of a customer buying latex paint given that they have bought a roller and paint is .45/.525 = .85714 I hope that explanation makes more sense to you. |
