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Seventy percent of people who hear worrisome noises in
their automobiles wait more than six weeks to visit a
mechanic. Of those people who waited, 32% found that
the original problem had caused another problem. Let \'s
name events W: Waiting more than six weeks, and P2: A
second problem was caused.
Referring to question 1, what is the probability that a
second problem was caused, if a person waited more than
six weeks to visit the mechanic? Enter your answer to
three decimal places (0.XXX).?
Referring to question 1, what is the probability that a
person waited too long to visit the mechanic, and a second
problem was caused? Enter your answer to three decimal
places (0.XXX).?
Solution
Prob of waiting more than 6 weeks to visit mechanic:
P(W)=.7
Of those people who waited, 32% found that the original
problem had caused another problem:
P(P2|W)=.32
prob of causing 2nd problem:
P(P2) = P(P2|W) * P(W) ............{from Bayes theorem}
P(P2) = .32*.7 = .224
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Seventy percent of people who hear worrisome noises in their automobiles wait more than six weeks to visit a mechanic. Of those people who waited, 32% found that the original problem had caused another problem. Let \'s name events W: Waiting more than six weeks, and P2: A second problem was caused. Referring to question 1, what is the probability that a second problem was caused, if a person waited more than six weeks to visit the mechanic? Enter your answer to three decimal places (0.XXX).? Referring to question 1, what is the probability that a person waited too long to visit the mechanic, and a second problem was caused? Enter your answer to three decimal places (0.XXX).? Solution Prob of waiting more than 6 weeks to visit mechanic: P(W)=.7 Of those people who waited, 32% found that the original problem had caused another problem: P(P2|W)=.32 prob of causing 2nd problem: P(P2) = P(P2|W) * P(W) ............{from Bayes theorem} P(P2) = .32*.7 = .224 Name: Description: ...
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