Question

: Consider two bags in which we have balls of three different colors. Details are in the following table.

Red Yellow Green Bag A 3 5 4 Bag B 2 4 x

A bag is chosen at random and then a ball is chosen.

A. If the probability of green ball is 21 5, find x. B. Find the probability of Bag A and it is given that the ball chosen is green.

Answer 1

solution:

Given data

Bag A contains

3 red , 5 yellow and 4 green balls

Total no.of balls in bag A = 3+5+4 = 12

Bag B contains

  2 red , 4 yellow and X green balls

Total no.of balls in bag B = 2+4+x = 6+x

a) Given that

when a bag is chosen at random and then a ball is drawn

Probability of green ball = P(G) = 2/5

Let A be the event of chosing Bag A and B be the event of chosing Bag B

P(A) = P(B) = 1/2

Then ,By total probability theorem

P(Green) = P(A)*P(G|A) + P(B)*P(G|B) = 2/5

(1/2 * 4/12 ) + (1/2 * x/x+6 ) = 2/5

1/6 + ( x / 2x+12 ) = 2/5

x / (2x+12) = 2/5 - 1/6

x / (2x+12) = 7/30

30x = 84+14x

16x = 84

x = 84/16

x = 5.25

x = 5

The no.of green balls in Bag B (X) = 5

b)

Probability of Bag A given the ball chosen is green = P(A|G)

= P(AG) / P(G)

= [ P(A) * P(G|A) ] / P(G) [ using Bayes theorem ]

= (1/2 * 4/12 ) / (2/5)

= (1/6) / (2/5)

= 5/12

Probability of chosing Bag A given the ball chosen is green = 5/12