A bag contains 20 coins. If the probability t | vedclass

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(D) Let $B_4$ be the event that there are $4$ biased coins and $B_5$ be the event that there are $5$ biased coins. Given $P(B_4) = 1/3$ and $P(B_5) =

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None of these

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(D) Let $B_4$ be the event that there are $4$ biased coins and $B_5$ be the event that there are $5$ biased coins. Given $P(B_4) = 1/3$ and $P(B_5) = 2/3$.To sort out all biased coins in exactly $10$ draws,the $10^{th}$ draw must be the last biased coin.For $B_4$: We need $3$ biased coins in the first $9$ draws and the $4^{th}$ biased coin on the $10^{th}$ draw. The probability is $\frac{1}{3} 	imes \frac{{}^4C_3 	imes {}^{16}C_6}{{}^{20}C_9} 	imes \frac{1}{11} = \frac{1}{3} 	imes \frac{4 	imes {}^{16}C_6}{{}^{20}C_9 	imes 11} = \frac{4}{33} \frac{{}^{16}C_6}{{}^{20}C_{10} 	imes (10/20)} = \frac{8}{33} \frac{{}^{16}C_6}{{}^{20}C_{10}}$.For $B_5$: We need $4$ biased coins in the first $9$ draws and the $5^{th}$ biased coin on the $10^{th}$ draw. The probability is $\frac{2}{3} 	imes \frac{{}^5C_4 	imes {}^{15}C_5}{{}^{20}C_9} 	imes \frac{1}{11} = \frac{2}{3} 	imes \frac{5 	imes {}^{15}C_5}{{}^{20}C_9 	imes 11} = \frac{10}{33} \frac{{}^{15}C_5}{{}^{20}C_{10} 	imes (10/20)} = \frac{20}{33} \frac{{}^{15}C_5}{{}^{20}C_{10}}$.Summing these,the probability is $\frac{2}{33} \left( \frac{4 \cdot {}^{16}C_6 + 10 \cdot {}^{15}C_5}{{}^{20}C_{10}} ight) = \frac{4}{33} \left( \frac{2 \cdot {}^{16}C_6 + 5 \cdot {}^{15}C_5}{{}^{20}C_{10}} ight)$.

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

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