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

• A. $\frac{241}{1456}$
• B. $\frac{164}{4165}$
• C. $\frac{451}{884}$
• D. None of these

Answer 1

The correct option is B

$\frac{164}{4165}$

Explanation for the correct option:

As we know, In a pack of cards there are four aces and $48$ other cards.

Step 1. Let event $\alpha$ be drawing $9$ cards which are not ace and $\beta$ be drawing an ace card.

Therefore, the required probability is

$P(\alpha \cap \beta )=P\left(\alpha \right)\times P\left(\beta \right)$

Step 2. There are four aces and $48$other cards.

$\therefore P\left(\alpha \right)=\frac{{}^{48}C9}{{}^{52}C9}$

Step 3. After drawing 9 non-ace cards, the 10th card must be ace.

$\therefore P\left(\beta \right)=\frac{{}^{4}C1}{{}^{42}C1}$

$=\frac{4}{42}$

Step 4. Put the value of $P\left(\alpha \right)$ and $P\left(\beta \right)$ in $P(A\cap B)$

$P(\alpha \cap \beta )=\frac{{}^{48}C9}{{}^{52}C9}.\frac{4}{42}$

$=\frac{48\times 47\times 46\times 45\times 44\times 43\times 42\times 41\times 40\times 39}{52\times 51\times 50\times 49\times 48\times 47\times 46\times 45\times 44\times 43}.\frac{4}{42}$

$=\frac{164}{4165}\mathbf{}$

Hence, option ‘B’ is Correct.