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In a city school of 1200 students 42% of the students are on the honorable 64% have a part time job and 28% are on the honor roll and have a part-time job what is the probability that a randomly selected student is on the honor roll given that the student has a part time job

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Answer:


P(H|P) = (P(H \cap P))/(P(P))

And replacing we got:


P(H|P) =(0.28)/(0.64)= 0.4375

And the probability required for this case is 0.4375

Explanation:

For this case we define the following events:

H = represent that the student is honorable

P= represent that the student have a part time job

And we have the following probabilities:


P(H) = 0.42 , P(P) = 0.64, P(H \cap P) =0.28

And we want to find this probability:


P(H|P)

And we can use the following probability:


P(H|P) = (P(H \cap P))/(P(P))

And replacing we got:


P(H|P) =(0.28)/(0.64)= 0.4375

And the probability required for this case is 0.4375

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