Question

A sports store owner conducts a game ‘weekend-surprise’ every Friday for his customers.

He fills two bags with cricket balls of red and white colours. The first bag has 4 white and 4 red balls, while the second bag contains 3 white and 5 red balls. The rules of the game are:

• The customer will be blind folded.
• Two balls have to be transferred from the first bag to the second bag one after another without replacement, and then one ball has to be drawn out from the second bag.
• The colours of the three balls (two balls transferred from the first bag and one ball drawn from the second bag) are considered.
• If all the three balls are of the same colour, the customer wins a surprise gift.

What is the probability that a customer can win the surprise gift?

Answer 1

Total Probability of Winning

E₁: Transferring 2 White balls from Bag 1 to Bag 2:

The probability of picking the first white ball is and the second is (since it’s without replacement).

P(W₁, W₂) = `4/8 × 3/ 7`

= `12/56`

`P(E_1)= 3/14`

E₂: Transferring 2 Red balls from Bag 1 to Bag 2:

P(E_2) = `(""^4C_2)/(""^8C_2) = 3/14`

E₃: One ball is Red and other is white 

P(E₃) = `(""^4C_1 × ""^4C_1)/(""^8C_2)`

= `4/7`

All three balls are of the same colour:

P(A) = P(E₁) × P(A/E₁) + P(E₂) × P(A/E₂)

= `3/14 × (""^5C_1)/(""^10C_1) + (3/14) × (""^7C_1)/(""^10C_1)`

P(A) = `9/35`