Question

Each of three bags A, B, C contains white balls and black balls. A has a1 white & b1 black, B has a2 white & b2 black and C has a3 white & b3 black. A ball is drawn at random and is found to be white. Find the respective probability that it is from A, B & C.

Answer 1

Answer:

See explanation ( Answers are too long)

Step-by-step explanation:

We will first compute a general probability for picking a white ball:

          P (W) = a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)

part a)

We are asked to find the probability of white ball given that it pulled from bag A. So if we express it in notation we are asked for P ( A / W). We will use conditional probability to answer our question:

                           P ( A / W ) = P ( W & A ) / P (W)

                           P ( W & A ) = a_1 / (a_1 + b_1)

Hence,

P ( A / W ) = [a_1 / (a_1 + b_1)] / [a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)]    

part b)

We are asked to find the probability of white ball given that it pulled from bag B. So if we express it in notation we are asked for P ( B / W). We will use conditional probability to answer our question:

                           P ( B / W ) = P ( W & B ) / P (W)

                           P ( W & B ) = a_2 / (a_2 + b_2)

Hence,

P ( A / W ) = [a_2 / (a_2 + b_2)] / [a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)]    

part c)

We are asked to find the probability of white ball given that it pulled from bag C. So if we express it in notation we are asked for P ( C / W). We will use conditional probability to answer our question:

                           P ( C / W ) = P ( W & C ) / P (W)

                           P ( W & C ) = a_3 / (a_3 + b_3)

Hence,

P ( A / W ) = [a_3 / (a_3 + b_3)] / [a_1 / (a_1 + b_1) + a_2 / (a_2 + b_2) + a_3 / (a_3 + b_3)]