Question 1134103
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The question asks for the probabilities of nearly all possible outcomes.  So you might as well get some practice calculating the probabilities by finding the probability of every possible outcome.  That is a good thing to do anyway, since you know you have made a calculation error if the sum of the probabilities is not 1.<br><pre>
(PPP): (0.8)(0.7)(0.7) = 0.392
(PPF): (0.8)(0.7)(0.3) = 0.168
(PFP): (0.8)(0.3)(0.2) = 0.048
(PFF): (0.8)(0.3)(0.8) = 0.192
(FPP): (0.2)(0.2)(0.7) = 0.028
(FPF): (0.2)(0.2)(0.3) = 0.012
(FFP): (0.2)(0.8)(0.2) = 0.032
(FFF): (0.2)(0.8)(0.8) = 0.128
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                  sum:   1.000<br></pre>
Now to answer the specific questions....<br>
(1) Find the probability that a worker will pass the first and the third test (presumably this means he does NOT pass the second....)
ANSWER: PFP = 0.048 = 4.8%<br>
(2) Find the probability that a worker will pass at least two tests
ANSWER: PPP  + PPF + PFP + FPP = 0.392+0.168+0.048+0.192 = 0.800 = 80%<br>
(3) Find the conditional probability that a worker will pass the first and the third test, given that a worker will pass at least two tests
ANSWER: PFP/(PPP + PPF + PFP + FPP) = 0.048/0.8 = .06 = 6% (approximately)<br>
Find the probability that a worker will fail all tests or pass all tests
ANSWERS: FFF = 0.128 = 12.8%; PPP = 0.392 = 39.2%