Cards are drawn one by one without replacement from a pack of 52 cards. The probability that 10 cares will precede the first ace is `241//1456` b. `18
Cards are drawn one by one without replacement from a pack of 52 cards.
The probability that 10 cares will precede the first ace is
`241//1456`
b. `18//625`
c. `451//884`
d. none of these
A. `241//1456`
B. `164//4168`
C. `451//884`
D. None of these
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Correct Answer - B
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`
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