Question
Mathematics Question on Multiplication Theorem on Probability
There are three bags B1,B2 and B3. The bag B1 contains 5 red and 5 green balls, B2 contains 3 red and 5 green balls, and B3 contains 5 red and 3 green balls. Bags B1,B2 and B3 have probabilities 103,103 and 104 respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
Probability that the chosen ball is green, given that the selected bag is B3, equals 83
Probability that the chosen ball is green equals 8039
Probability that the selected bag is B3, given that the chosen ball is green, equals 135
Probability that the selected bag is B3 and the chosen ball is green equals 103
Probability that the chosen ball is green equals 8039
Solution
P(B1)=103P(B2)=103P(B3)=104 (A) Probability that selected bag is B3 and the chosen ball is green =P(B3)×P(B3G) =104×83=203 (B) Probability that the selected bag isB3 given that the chosen ball is green P(GB3) P(GB3)=P(B1G)P(B1)+P(B2G)P(B2)+P(B3G)P(B3)P(B3G)P(B3) =103×105+103×85+104×83104×83=134 (C) Probability that the chosen ball is green, given that the selected bag is B3 P(B3G)=83 (D) Probability that the chosen ball is green P(G)=P(B1)P(B1G)+P(B2)P(B2G)+P(B3)P(B3G) =103×105+103×85+104×83 =8039