Question

A bag contains 4 yellow and 5 red balls and another bag contains 6
yellow and 3 red balls. A ball is drawn from the first bag and without
seeing its colour, it is put into the second bag. Find the probability
that if now a ball is drawn from, the second bag, it is yellow in colour.​

Answer 1

Answer:

Bag 1 contains 5 red and 3 green balls.

Bag 2 contains 4 red and 6 green balls.

Probability = No.of favorable outcomes/Total No.of outcomes

Bag 1 contains 8 balls

Probability of selecting a red ball from bag 1

P(R1) = 5C1(5 red balls)/8C1 = 5/8

Probability of selecting a green ball from bag 1

P(G1) = 3C1(3 green balls)/8C1 = 3/8

Bag 2 contains 10 balls.

Probability of selecting a green ball from bag 2

P(G2) = 6C1(6 green balls)/10C1 = 6/10 = 3/5

Probability of selecting a red ball from bag 2

P(R2) = 4C1(4 red balls)/10C1 = 4/10 = 2/5

We have 2 possibilities here,

1. Probability of drawing a red ball from bag 1 and green ball from bag 2

(PT1) = P(R1) x P(G2)

= 5/8 x 3/5

= 3/8

We are multiplying here because, the selection of red balls does't depend on green balls and vice versa,i.e both are independent.

2. Probability of drawing a green ball from bag 1 and red ball from bag 2

P(T2) = P(R2) x P(G1)

= 2/5 x 3/8

= 6/40

To get the overall probability, we take the sum of individual probabilities.

Total probability = P(T1) + P(T2)

= 3/8 + 6/40

= 21/40