Question

Bag 1 contains 3 red balls and 7 green balls. Bag 2 contains 8 red balls and 4 green balls. Bag 3 contains 5 red balls and 11 green balls. Bag 1 is chosen twice as often as either Bag 2 or Bag 3 (which have the same probability of being chosen). After a bag is chosen, a ball is chosen at random from that bag. Calculate the probability that:

a) a red ball is chosen

b) a red ball from Bag 2 is chosen

c) if it is known that a red ball is chosen, what is the probability that it comes from Bag 2?

Answer 1

B1 : Event of Choosing Bag 1

B2 : Event of choosing Bag 2

B3 : Event of choosing Bag 3

Bag 1 is chosen twice as often as either Bag 2 or Bag 3 i.e

2P(B1) = P(B2) = P(B3)

P(B1)=1/2 ; P(B2)=P(B3) =1/4

a)

R :Event of red ball is chosen

P(R|B1) =

Probability chooing Red ball given that Bag 1 was chosen

= Number of red balls in Bag1 /Total number of balls in Bag1 = 3/10

P(R|B2) =

Probability chooing Red ball given that Bag 2 was chosen

= Number of red balls in Bag2 /Total number of balls in Bag2 = 8/12

P(R|B3) =

Probability chooing Red ball given that Bag 3 was chosen

= Number of red balls in Bag3 /Total number of balls in Bag3 = 5/16

probability that  a red ball is chosen = P(R)

P(R) = P(B1)P(R|B1)+P(B2)P(R|B2)+P(B3)P(R|B3)

= (1/2) x (3/10) + (1/4) x (8/12) + (1/4)x(5/16) = 3/20 + 8/48 + 5/64 = 3/20+1/6+5/64 = (379/960)=0.394791667

Probability that  a red ball is chosen = 379/960 = 0.394791667

b)

Probability that a red ball from Bag 2 is chosen = P(B2 and R) = P(B2) P(R|B2) = (1/4)x(8/12) =1/6 = 0.166667

Probability that a red ball from Bag 2 is chosen = 1/6 = 0.166667

c)

If it is known that a red ball is chosen, Probability that it comes from Bag 2= P(B2|R)

From (b)

P(B2 and R) = P(B2) P(R|B2) = 1/6

From (a) P(R) = P(B1)P(R|B1)+P(B2)P(R|B2)+P(B3)P(R|B3) = 379/960

If it is known that a red ball is chosen, Probability that it comes from Bag 2= 160/379 = 0.422163588