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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 theseA. `241//1456`B. `164//4168`C. `451//884`D. None of these |
Answer» 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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