Answer to Question #130221 in Statistics and Probability for Talisha

Question

Sony produces its TV sets in 3 manufacture plants A, B & C.
 Plant A produces 50% of the TV sets & the probability that a TV set
manufactured here is defective is 0.02.
 Plant B produces 30% of the TV sets & the probability of defective is 0.05.
 Plant C produces 20% of the TV sets & the probability that a TV set
manufactured here is defective is 0.01.
a) Make the probability Tree of this situation.
b) If a Sony TV is randomly selected. What is the probability that it is
defective?
c) If a TV is selected at random & found to be defective then what is the
probability that it was manufactured in plant B?

Answer 1

ANSWER A

PROBABILITY TREE IS AS BELOW

ANSWER B

PROBABILITY OF A RANDOM SELECTION OF A SONY TV TO BE DEFECTIVE IS GIVEN BY,

P(d) = P(a)*P(ad) + P(b)P(bd) + P(c)P(cd)

P(d) = 0.5*0.02 + 0.3*0.05 + 0.2*0.01

P(d) = 0.027

ANSWER C

We have to find the prbability that the randomly selected (defective) TV is from plant B given that it is defective = P(TV is from plant B | TV is defective)

We know that $P(A|B) = \frac{P(A \bigcap B)}{P(B)}$

Thus we have to find $\frac{P(TV\ is\ from\ plant B\ and\ is\ defective)}{P(TV is \ defective)} = \frac{0.3*0.05}{0.027} = 0.5556$