From Book Basic Propability and Statistics for Engineers andScientists 3 of Hayt

Question

From Book Basic Propability and Statistics for Engineers andScientists 3 of Hayter.

1.9.11 Bar1 contains 6 red balls, 7 blue balls , and 3 green balls.Bag 2 contains 8 red balls, 8 blue balls and 2 green balls. Bag 3contains 2 red balls, 9 blue balls, and 8 green balls. Bag 4contains 4 red balls, 7 blue balls, and no green balls. Bag 1 ischosen with a probability 0.15, bag 2 with a probability of 0.20,bag 3 with a probability of 0.35 and bag 4 with a probability of0.30, and then a ball is chosen at random from the bag. Calculate:the probabilities that that a blue ball is chosen, theprobability that ball 4 was chosen if the ball is green andthe probability that bag 1 was chosen if the ball is blue.

Explanation / Answer

a)P(B)=P(Bnb1)+P(Bnb2)+P(Bnb3)+P(Bnb4) Now each term P(Bnb)=P(B/b)P(b) so we have P(B)=(7/16)(.15)+(8/18)(.2)+(9/19)(.35)+(7/11)(.3)=.511 b)P(b4/G)=0 c)P(b1/B)=P(b1nB)/P(B)=P(B/b1)P(b1)/P(B) So P(b1/B)=(7/16)(.15)/.511=.128

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