A bag contains 20 coins. If the probability that the 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 bag is exactly 10 draws is
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

Options

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

Answer

C

Explanation

P(4 biased coins `=1/3`
P(5 biased coins) `=1/4`
The required proability is
`1/3(""^(4)C_(3)""^(16)C_(3))/(""^(20)C_(9))(1)/(""^(11)C_(1))+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))]`