A company manufacturing electronic components for home entertainment systems buy

Question

A company manufacturing electronic components for home entertainment systems buys electrical connectors from three suppliers. It buys 20% of the connectors from supplier A, 33% from supplier B, and 47% from supplier C. It finds that 3% of the connectors from supplier A are defective. 2% of the connectors from supplier B are defective, and 5% of the connectors from supplier C are defective. If a customer buys one of these components and finds that the connector is defective, what is the probability that it came from supplier B? A game is said to be "fair" if the expected value for winnings is 0, that is, in the long run, the player can expect to win 0. Consider the following game. The game costs $1 to play and the winnings are $5 for red. $3 for blue, $2 for yellow, and nothing for white. The following probabilities apply. What are your expected winnings? Does the game favor the player or the owner? A gambler claimed that he had loaded a die so that it would hardly ever come up 1. He said the outcomes of 1, 2,3, 4, 5, 6 would have probabilities 1/20, 1/6, 1/6, 1/6,1/6, 16, respectively. Can he do what he claimed? Why or why not? Is a probability distribution described by listing the outcomes along with their corresponding probabilities? Why or why not?

Explanation / Answer

47)

A game is said to be "fair" if the expected value for winnings is 0, that is, in the long run, the player can expect to win (or lose) $0.

Consider the following game. The game costs $1 to play and the winnings are $5 for red, $3 for blue, $2 for yellow and nothing for white. The following probabilities applyl. What are your expected winnings? Does the game favor the player or the owner?
Outcome Probability
Red .02
Blue .04
Yellow .16
White .78

E(x) = 0.02*4 + 0.04*2 + 0.16*1 + 0.78*(-1)
= -0.46

Game favors the owner.

46) Let the three components are A, B and c.

D be the defective.

Given that P(A) = 20% = 0.2

P(B) = 33% = 0.33

P(C) = 47% = 0.47

P(D/A) = 3% = 0.03

P(D/B) = 2% = 0.02

P(D/C) = 5% = 0.05

W and we have to find P(B/D).

P(B/D) = P(D/B)*P(B) / P(D/A)P(A) + P(D/B)P(B) + P(D/C)P(C)

= 0.02*0.33 / (0.03*0.2 + 0.02*0.33 + 0.05*0.47)

= 0.0066 / 0.006 + 0.0066 + 0.0235

= 0.0066 / 0.0361

= 0.1828

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