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A system is given in figure below (lines symbolizes connection between operation

ID: 3297640 • Letter: A

Question

A system is given in figure below (lines symbolizes connection between operations). Assume the operations fail independently. If one fails the system stop The probability of that particular component working is in parenthesis. What is the probability that the system is working? Materials In _____ Operation A(0.90)_____Operation B(0.75)_Operation C(0.85)_____Operation D(0.80) _____Operation E(0.80)_____ Finished products out a. 0.50 b. 0.4131 c. 0.80 d. 0.3672 A consumer electronic chain store sells three different brands of TVs or its TV sales 35% are brand 1 (the least expensive), 35% are brand 2 and 30% are brand 3. Each manufacturer offers 1-year warranty on parts and labor. It is known that 25% of brand 1's TVs require warranty repair work, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively. If a customer returns to the store with a TV that needs warranty repair work (this is given it needs repair work), what is the probability that it is a brand 1 TV? a. 0.359 b. 0.195 c. 0.467

Explanation / Answer

Question 5:

Probability that the system is working = Probability that each component is working and as they all are independent , this could be computed as: ( using the multiplication rule for independent events )

= 0.9*0.75*0.85*0.8*0.8 = 0.3672

Therefore d) 0.3672 is the required probability here.

Question 6:

Here the sales data we are given that:

P( Brand 1) = 0.35, P( brand 2) = 0.35 and P( Brand 3) = 0.3

P( W | Brand 1) = 0.25 that is 25% of brand 1 TVs require warranty repair work

Similarly, P( W | Brand 2) = 0.2 and P( W| Brand 3) = 0.1

P(W) = P( W | Brand 1) P( Brand 1) + P( W | Brand 2) P( Brand 2) + P( W | Brand 3) P( Brand 3)

P(W) = 0.25*0.35 + 0.2*0.35 + 0.1*0.3 = 0.1875

Now given that there is a repair work to be done, probability that it was a brand 1 TV would be computed as:

= P( W | Brand 1) P( Brand 1) / P(W)

= 0.25*0.35 / 0.1875

= 0.467

Therefore 0.467 is the required probability here.

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