First slide

Theorems of probability

Question

A bag has 10 balls. Six balls 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

Moderate

A

$\frac{5}{21}$
B

$\frac{2}{21}$
C

$\frac{7}{21}$
D

$\frac{3}{21}$

Solution

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 $\frac{^{6} {C_{2}}^{4} C_{3}}{^{10} C_{5}} = \frac{5}{21}$

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