#### Question

A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.

#### Solution

Bag X = 4 white, 2 black.

Bag Y = 3 white, 3 black

Let A be the event of selecting one white and one black ball.

E_{1} first bag selected

E_{2} second bag selected

`P(E_1)=1/2 P(E_2)=1/2`

`P(A/E_1)=4/6xx2/5+2/6xx4/5=16/30`

`P(AE_2)=3/6xx3/5+3/6xx3/6xx3/5=18/30`

`P(E_2"/"A)=(P(E_2)P(A"/"E_2))/(P(E_1)P(A"/"E_1)+P(E_2)P(A"/"E_2)`

`P(E_2"/"A)= (1/2xx18/30)/(1/2xx16/30+1/2xx18/30)`

`=18/(16+18)`

=`18/34`

=`9/17`

Is there an error in this question or solution?

Solution A Bag X Contains 4 White Balls and 2 Black Balls, While Another Bag Y Contains 3 White Balls and 3 Black Balls. Two Balls Are Drawn (Without Replacement) at Random from One of the Bags and Were Found to Be One White and One Black. Concept: Conditional Probability.