Consider 3 urns. Urn A contains 2 white and 4 red balls, urnBcontains 8 white and 4 red balls, and urnCcontains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ballchosen from urnAwas white given that exactly 2 white balls were selected?

P ( A / E ) = 0.6365

Step-by-step explanation:

Given:

- Urn A : 2 White , 4 Red

- Urn B : 8 White , 4 Red

- Urn C : 1 White , 3 Red

- White = W , Red = R

Find

If 1 ball is selected from each urn, what is the probability that the ball chosen from urn Awas white given that exactly 2 white balls were selected?

Solution:

- Let event E be selecting exactly 2 white balls after going through each urn.

- The possible outcomes for event E can be expressed as:

                       { ( W , W , R ) : ( W , R , W ) : ( R , W , W ) }

- The probability of event E would be as such:

                 P ( E ) = P ( W , W , R ) + P( W , R , W ) + P ( R , W , W )

                 P ( E ) = ( 2/6 * 8/12 * 3/4 ) + ( 2/6 * 4/12 * 1/4 ) + ( 4/6 * 8/12 * 1/4 )

                 P ( E ) = 0.1667 + 0.0278 +  0.11111

                 P ( E ) = 0.30561

- Let event A be selecting a white ball from urn A and exactly two white balls in total are picked:

                 P ( A & E ) = P ( W , W , R ) + P( W , R , W )

                 P ( A & E ) = ( 2/6 * 8/12 * 3/4 ) + ( 2/6 * 4/12 * 1/4 )

                 P ( A & E ) = 0.1667 + 0.0278

                 P ( A & E ) = 0.1944

- The conditional probability of choosing a white ball from urn A given that exactly 2 W balls are selected among three urns:

                P ( A / E ) = P ( A & E ) / P ( E )

                P ( A / E ) = 0.1944 / 0.30561

                P ( A / E ) = 0.6365

answer: the answer is ann