A bag contains 20 coins. If the probability that the bag contains exactly 4 biased coin is 3/4 and that of exactly 5 biased coin is 2/3, then the probability that all the biased coin are sorted out from bag is exactly 10 draws is `5/(10)(^(16)C_6)/(^(20)C_9)+1/(11)(^(15)C_5)/(^(20)C_9)` b. `2/(33)[(^(16)C_6+5^(15)C_5)/(^(20)C_9)]` c. `5/(33)(^(16)C_7)/(^(20)C_9)+1/(11)(^(15)C_6)/(^(20)C_9)` d. none of these

Options

A. `5/10(""^(16)C_(6))/(""^(20)C_(9))+1/11(""^(15)C_(5))/(""^(20)C_(9))`
B. `2/(33)[(""^(16)C_6+5^(15)C_5)/(""^(20)C_9)]`
C. `5/33(""^(16)C_(7))/(""^(20)C_(9))+1/11(""^(15)C_(6))/(""^(20)C_(9))`
D. None of these

Answer

B

Explanation

P (4 bases coin)`=1/3`
P(5 biased coin) `=1/4`
Hence, the required probability is
`=1/3(""^(4)C_(3)""^(16)C_(6))/(""^(20)C_(9))+2/3(""^(5)C_(4)""^(15)C_(5))/(""^(20)C_(9))(1)/(""^(11)C_(1))`
`=2/33[(""^(16)C_(6)+5""^(15)C_(5))/(""^(20)C_(9))]`