Probability Question?
1. he had no safety instruction
2. he had safety instruction
Answer:
Dear mj D,
If you adopt the relative frequencies listed in the problem as your probabilities, then you get the following, starting by writing what you already know.
P(A) = 0.01 is the probability of an EEE having an accident during the year, so
P(~A) = 1 - P(A) = 0.99 is the probability of an EEE being accident-free during the year.
P(S | A) = 0.40 is the probability that an EEE has safety instructions, given the EEE has an accident during the year, so
P(~S | A) = 1 - P(S | A) = 0.60 is the probability that an EEE has no safety instructions, given the EEE has an accident during the year.
P(~S) = 0.9 is the probability that an EEE has no safety instructions, so
P(S) = 1 - P(~S) = 0.1 is the probability that an EEE has safety instructions.
1. Now calculate the probability of an EEE being accident-free during the year, given the EEE has no safety instructions.
P(~A | ~S) = P(~A & ~S) / P(~S)
= [P(~S) - P(A & ~S)] / P(~S)
= [P(~S) - P(A) P(~S | A)] / P(~S)
= [(0.9) - (0.01) (0.60)] / (0.9)
= [(0.9) - (0.006)] / (0.9)
= [0.894] / (0.9)
= 149 / 150
= 0.9933 (to four decimal places).
2. Similarly, you can calculate the probability of an EEE being accident-free during the year, given the EEE has safety instructions.
P(~A | S) = P(~A & S) / P(S)
= [P(S) - P(A & S)] / P(S)
= [P(S) - P(A) P(S | A)] / P(S)
= [(0.1) - (0.01) (0.4)] / (0.1)
= [(0.1) - (0.004)] / (0.1)
= [0.096] / (0.1)
= 24 / 25
= 0.96 .
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