**Question **: A bag contains 20 coins. If the probability that the bag contains exactly 4 biased coin is 1/3 and that of exactly 5 biased coin is 2/3, then the probability that all the biased coin are sorted out from the bag in exactly 10 draws is

(A) \dfrac{5}{10}\cdot \dfrac{16C_{6}}{20C_{9}}+\dfrac{1}{10}\cdot \dfrac{15C_{5}}{20C_{9}}

(B) \dfrac{2}{33}\left[ \dfrac{16C_{6}+\left( 5\right) ^{15}C_{5}}{20C_{9}}\right]

(C) \dfrac{5}{33}\cdot \dfrac{16C_{6}}{20C_{9}}+\dfrac{1}{11}\cdot \dfrac{15C_{5}}{20C_{9}}

(D) none of these

**Answer **: option (B)

**Solution **:

P(4 biased coin) = 1/3

P(5 biased coin) = 1/4

Hence, the required probability is

=\begin{aligned}\cdot \ \dfrac{1}{3}\cdot \dfrac{4C_{3}\cdot 16C_{6}}{20C_{9}}+\dfrac{2}{3}\cdot \dfrac{5C_{4}.15C_{5}}{20C_{9}}\cdot \dfrac{1}{11C_{1}}\end{aligned}

=\dfrac{2}{33}\left[ \dfrac{16C_{6}+\left( 5\right) ^{15}C_{5}}{20C_{9}}\right]