14.Probability
normal

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

Solution

$\mathrm{P}(4 \text { biased coin })=\frac{1}{3} ; \mathrm{P}(5 \text { biased coin })=\frac{1}{4}$ Required probability

$ = \frac{1}{3}\frac{{{\,^4}{C_3}^{16}{C_6}}}{{{\,^{20}}{C_9}}} = \frac{1}{{{\,^{11}}{C_1}}} + \frac{2}{3}\frac{{{\,^5}{C_4}^{15}{C_5}}}{{{\,^{20}}{C_9}}} = \frac{1}{{{\,^{11}}{C_1}}}$

$ = \frac{2}{{33}}\left[ {\frac{{2 \cdot {\,^{16}}{C_6} + 5 \cdot {\,^{15}}{C_5}}}{{^{20}{C_9}}}} \right]$

Standard 11
Mathematics

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