First slide

Theorems of probability

Question

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

Moderate

A

$\frac{214}{1456}$
B

$\frac{164}{4165}$
C

$\frac{451}{884}$
D

None of these

Solution

Let event Abe drawing cards which are not ace and B be drawing an ace card. Therefore, the required probability is
$P (A \cap B) = P (A) \times P (B)$
Now, there are four aces and 48 other cards. Hence,
$P (A) = \frac{^{48} C_{9}}{^{52} C_{9}}$
After having drawn 10 non-ace cards, the l1th card must be ace.
Hence,
$P (B) = \frac{^{4} C_{1}}{^{42} C_{1}} = \frac{4}{42}$
Hence,
$P (A \cap B) = \frac{^{48} C_{9}}{^{52} C_{9}} \frac{4}{42}$

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