If B) =-, P( AIB)=-, P(AUB)=-, what are P (B), P (A) and P (BIA)? P(Ar [Bayes\'

Question

If B) =-, P( AIB)=-, P(AUB)=-, what are P (B), P (A) and P (BIA)? P(Ar [Bayes' Rule, 15 points] 40% of the sales at a car dealership are new cars, the rest are used, within the first year, one-fifth of new cars and one-third of the used cars need repairs covered by the warranty. If a car needs warranty-covered work, what is the probability it is a new car? 5. 6. (Sample Spaces, 12 points] Given 4 cards labeled K,Q.J,A we keep turning over cards (without replacement) until we see the card labeled A. List the sample space. You may use a tree.

Explanation / Answer

Question 5:

Here we are given that:

P( new cars ) = 0.4 and P( used cars ) = 1 - 0.4 = 0.6

Also we are given that: one - fifth of the new cards and one - third of the used cars need repairs . Therefore

P( repairs | new cars ) = 0.2 and P( repairs | used cars ) = 1/3 = 0.3333

Using law of total probability we get:

P( repairs ) = P( repairs | new cars )P( new cars ) + P( repairs | used cars ) P( used cars )

P( repairs ) = 0.2*0.4 + (1/3)*0.6 = 0.28

Now given that a car needs a repair, probability that its a new car is computed as ( Using Bayes theorem )

P( new cars | repairs ) = P( repairs | new cars )P( new cars ) / P( repairs ) = 0.2*0.4 / 0.28 = 0.2857

Therefore 0.2857 is the required probability here.

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