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