A bag has 10 balls. Six ball are drawn in an attempt and replaced. Then another draw of 5 balls is made from the bag. The probability that exactly two balls are common to both the draw is `5//21` b. `2//21` c. `7//21` d. `3//21`

Options

A. `5//21`
B. `2//21`
C. `7//21`
D. `3//21`

Answer

A

Explanation

The total number of ways of making the second draw is `.^(10)C_(5)`.
The number of draws of 5 balls containing 2 balls common with first draw of 6 balls is `.^(6)C_(2).^(4)C_(3)`. Therefore, the probability is
`(.^(6)C_(2).^(4)C_(3))/(.^(10)C_(5) )= (5)/(21)`