My Question is as following: Forty-two percent of a corporation’s blue-collar em

Question

My Question is as following:

Forty-two percent of a corporation’s blue-collar employees were in favor of a modified health care plan, and 22% of its blue-collar employees favored a proposal to change the work schedule. Thirty-four percent of those favoring the health care plan modifi- cation favored the work schedule change.

a) What is the probability that a randomly selected blue-collar employee is in favor of both the modified health care plan and the changed work schedule?

b) What is the probability that a randomly chosen blue-collar employee is in favor of at least one of the two changes?

c) What is the probability that a blue-collar employee favoring the work schedule change also favors the modified health care plan?

Thank you!! Greeting

I posted this question yesterday, but the answer i got then was absolutely wrong. Would appriciate som real help here!

Explanation / Answer

P( Favour modified health care plan) = P(A) = 42% = 0.42

P ( Favour to change the work schedule) = P(B) = 22% = 0.22

P ( B /A) = 0.34

(a) probability that a randomly selected blue-collar employee is in favor of both the modified health care plan and the changed work schedule?

P ( A and B) = P ( B /A) * P(A) = 0.34 * 0.42 = 0.1428

(b) What is the probability that a randomly chosen blue-collar employee is in favor of at least one of the two changes?

P( A or B) = P(A) + P(B) - P(A and B) = 0.42 + 0.22 - 0.1428 = 0.4972

(c) What is the probability that a blue-collar employee favoring the work schedule change also favors the modified health care plan?

P(A/B) = P ( B /A) * P(A) / ( P ( B /A) * P(A) + P(B/A') P(A')) [ where P(A') = 1 - P(A) = 0.58

P(A/B) = (0.34 * 0.42)/ (0.34 * 0.42 + (0.22 - 0.34 * 0.42) )

           = 0.1428 / 0.22 = 0.6491

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