ProbabilityHard
Question
Box-I contains 5 red and 4 blue balls while box-II contains 4 red and 2 blue balls. A fair die is thrown. If it turns up a multiple of 3, a ball is drawn from box-I, else a ball is drawn from box-II. Find the probabiliy of the event ′ball drawn is from box-I, if it is blue′.
Options
A.1/6
B.15/19
C.4/19
D.2/5
Solution
A → Ball from Box I, P(A) = 
P(E/A) = 4/9
B → Ball from Box II P(B) =
P(E/B) = 2/6
E → Ball drawn is Blue
By use of Bays theorem
P(A/E) =
P(E/A) = 4/9B → Ball from Box II P(B) =
P(E/B) = 2/6E → Ball drawn is Blue
By use of Bays theorem
P(A/E) =
Create a free account to view solution
View Solution FreeMore Probability Questions
Three athlete A,B and C participate in a race competition. The probability of winning A and winning of B is twice of win...If P(B) = , P(A ∩ B ∩ )= and P( B ∩ = , then P(B ∩ C) is...There are n letters and n addressed envelopes. The probability that all the letters are not kept in the right envelope, ...For three events A, B and C, P (exactly one of the events A or B occurs) = P (exactly one of the events B or C occurs) =...The probability that A speaks truth is , while this probability for B is . The probability that they contradict each oth...