SOLUTION: In a certain year, the probability that a person in the United States would declare personal bankruptcy was 0.004. The probability that a person in the United States would declare

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Question 1196442: In a certain year, the probability that a person in the United States would declare personal bankruptcy was 0.004. The probability that a person in the United States would declare personal bankruptcy and had recently experienced a "big three" event (loss of job, medical problem, or divorce or separation) was 0.002. What was the probability that a person had recently experienced one of the "big three" events, given that she had declared personal bankruptcy? (Round your answer to four decimal places.)
Answer by ikleyn(52835) About Me  (Show Source):
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In a certain year, the probability that a person in the United States would declare personal bankruptcy was 0.004.
The probability that a person in the United States would declare personal bankruptcy and had recently
experienced a "big three" event (loss of job, medical problem, or divorce or separation) was 0.002.
What was the probability that a person had recently experienced one of the "big three" events,
given that she had declared personal bankruptcy? (Round your answer to four decimal places.)
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Let A be an event that a person in the United States would declare personal bankruptcy.


Let B be an event that that a person in the United States had recently experienced a "big three" event 
(loss of job, medical problem, or divorce or separation).


The problem wants you find a CONDITIONAL PROBABILITY of the event P(A and B) given A.


This probability is  the ratio

    P(A and B)/P(A) = 0.002%2F0.004 = 1%2F2 = 0.5 = 50%.    ANSWER

Solved and thoroughly explained.