SOLUTION: The workers in a VS shop need to attend a competency course and pass three tests. The probability of passing the first test is 0.8 and if a worker passes a test, the probability th

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Question 1134103: The workers in a VS shop need to attend a competency course and pass three tests. The probability of passing the first test is 0.8 and if a worker passes a test, the probability that the worker will pass the subsequent test is 0.7. Instead, if the worker fails, the probability that the worker will fail the subsequent test is 0.8. Find the probability that a worker will pass the first and the third test? Find the probability that a worker will pass at least two tests?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? Find the probability that a worker will fail all tests or pass all tests?
Found 2 solutions by Shin123, greenestamps:
Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
There is a 0.8 chance of passing the first test.

Case 1: Test 2 is failed.


If Test 2 is failed, there is a (1-0.8)=0.2 chance of success.

Case 2: Test 2 is passed.


If Test 2 is passed, there is a 0.7 chance of success. So there is a (0.8*0.2)+(0.8*0.7)=0.72 chance of passing the first and third test. There is a 0.8*0.7*0.7=0.392 chance of passing all 3 tests. There is a (1-0.8)*0.8*0.8=0.128 chance of failing all 3 tests. Figure out the rest yourself. Bonus: click this surprise link.

Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


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.
(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
                     ----------
                  sum:   1.000

Now to answer the specific questions....

(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%

(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%

(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)

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%