PART A AND B Three machines M1, M2, and M3 produce 40%, 45%, and 15%, respective
ID: 3329061 • Letter: P
Question
PART A AND B
Three machines M1, M2, and M3 produce 40%, 45%, and 15%, respectively, of the total number of nuts produced by a certain factory. The percentages of defective output of these machines are 3%, 6%, and 9%. A) If a nut is selected at random, find the probability that the item is defective. B) Suppose that a nut is selected at random and is found to be defective. Find the probability that the defective nut was produced by machine M1 M1-produced by machine 1 M2 = produced by machine 2 M3 = produced by machine 3 D = nut is defective D' = nut is not defective Let: Which of the following is the correct representation of the given probability of 9%? P(M3 nD) 0.09 P(D | M3 ) = 0.09 , p(M3 | D ) = 0.09 (O) P(D n M3 ) = 0.09 The route used by a certain motorist in commuting to work contains two intersections with traffic signal lights. The probability that she must stop at the first signal and second signal are 0.40 and 0.50, respectively. The probability that she must stop at either signal is 0.60. What is the probability that she must stop at the first signal but not the second signal F = must stop at first signal F' do not have to stop at first signal s must stop at second signal S' = do not have to stop at second signal Let: Match the correct probability description to each given probability value 0.40 0.50 0.50 Find A. P(F I S') B. P(S) C. P(F n S) D. P(S F) E. P(F) G. P(F US)Explanation / Answer
1) A) P(defective) = 0.4 * 0.03 + 0.45 * 0.06 + 0.15 * 0.09 = 0.0525
B) P(defective nut produced by machine M1) = 0.4 * 0.03 / 0.0525 = 0.012/0.0525 = 0.22857
C) Option-B is the correct answer.
2) P(F and S) = 0.40 + 0.50 - 0.60 = 0.3
P(F and S') = P(F) - P(F and S') = 0.40 - 0.3 = 0.1
A) P(F | S') = P(F and S') / P(S')
= 0.1/0.5 = 0.2
B) P(s) = 0.5
C) P(F and S) = 0.3
D) P(S' | F) = P(S' andF) / P(F)
= 0.1/0.4 = 0.25
E) P(F) = 0.4
F) P(F and S') = 0.1
G) P(F U S) = 0.6
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