Cards are drawn one by one without replacement from a pack of 52 cards. The probability that 10 cards will precede the first ace is

Options

A. A. `241//1456`
B. B. `164//4168`
C. C. `451//884`
D. D. None of these

Answer

B

Explanation

Let even A be drawing 9 cards which are not ace and B ve drawing an ace card. Therefore, the required probability is `P(AnnB)=P(A)xxP(B)`
Now, there are four aces and 48 other cards. Hence,
`P(A)=(""^(48)C_(9))/(""^(52)C_(9))`
After having drawn 9 non-ace cards, the `10^(th)` card must be ace. Hence,
`P(B)=(""^(4)C_(1))/(""^(42)C_(1))=4/42`
Hence, `P(AnnB)=(""^(48)C_(9))/(""^(52)C_(9))4/42`