Question

Box A contains 6 red balls and 3 green balls, whereas box B contains 3 red ball and 15 green balls.

Stage one. One box is selected at random in such a way that box A is selected with probability 1/5 and box B is selected with probability 4/5.

Stage two. Finally, suppose that two balls are selected at random with replacement from the box selected at stage one.

g) What is the probability that both balls are red?

h) Given that both balls are red, what is the probability they came from box A?

i) What is the probability that one ball is red and the other is green?

j) Given that one ball is red and the other is green, what is the probability they came from box A?

Answer 1

P(A) = P(box A is selected) =1/5 ,

P(B) = P( box B is selected) = 4/5

Selection with replacement is being done

g)

P(both balls are red) =

P( both are red balls given that box A was selected) × P(A)

+

P( both are red balls given that box B was selected) × P(B)

= (6/9)×(6/9) ×(1/5) + (3/15)×(3/15)×(4/5)

= 161/1125 = 0.1431

h)

It is a case of Bayes theorem,

P(box A was selected given that both balls were red)

= P( both balls are red given box A was selected) × P( box A is selected)  / P( both balls are red)

=( 6/9) ×(6/9) × ( 1/5) / (161/1125)

= 100/161 = 0.6211

i) 4 possible cases are

Bag A : (red & green) , ( green & red)

Bag B: (red & green) , ( green & red)

P( one is red and other is green)

=( 1/5) ×(6/9)×(3/9)×2 + (4/5)×(3/18)×(15/18)×2

= 14/45 = 0.3111

j) P(box A is selected given that one was red and other was green )

= P(one is red and other one is green given that box A is selected) × ( box A is selected)

/

P( one is red and other is green)

= ((6/9)×(3/9))×(1/5) / ( 14/45)

= 1/7 = 0.1428