A bag contains $20$ coins. If the probability that 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 bag contains $20$ coins. If the probability that 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
$\frac{5}{{33}}\frac{{{}^{16}{C_6}}}{{{}^{20}{C_9}}} + \frac{1}{{11}}\frac{{{}^{15}{C_5}}}{{{}^{20}{C_9}}}$
- B
$\frac{2}{{33}}\left( {\frac{{2.{}^{16}{C_6} + 5{}^{15}{C_5}}}{{{}^{20}{C_9}}}} \right)$
- C
$\frac{2}{{33}}\frac{{{}^{16}{C_7}}}{{{}^{20}{C_9}}} + \frac{1}{{11}}\frac{{{}^{15}{C_6}}}{{{}^{20}{C_9}}}$
- D
none of these
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