Question

From a pack of 52 playing cards; half of the cards are randomly removed without looking at them. From the remaining cards, 3 cards are drawn randomly. The probability that all are king.

• A. $\frac{1}{(25)(17)(13)}$
• B. $\frac{1}{(25)(15)(13)}$
• C. $\frac{1}{(52)(17)(13)}$
• D. $\frac{1}{(25)(51)(13)}$

Answer 1

Answer: (A)

$\frac{1}{(25)(17)(13)}$

Answer 2

Let $N=52$ be the total number of cards in a standard deck. The number of kings is $K=4$.
Half of the cards are randomly removed, so $26$ cards are removed, and $26$ cards remain.
Let $n=26$ be the number of remaining cards.
From these $n=26$ cards, 3 cards are drawn randomly. We want to find the probability that all 3 drawn cards are kings.

Let $X$ be the number of kings among the 26 cards that are removed. The number of kings among the remaining 26 cards is $4-X$.
The number of ways to choose 26 cards to remove from the 52 cards is $\binom{52}{26}$.
The number of ways to choose $X$ kings from the 4 kings and $26-X$ non-kings from the 48 non-kings is $\binom{4}{X} \binom{48}{26-X}$.
The probability that exactly $X$ kings are removed is $P(X) = \frac{\binom{4}{X} \binom{48}{26-X}}{\binom{52}{26}}$.
The possible values for $X$ are $0, 1, 2, 3, 4$.

Given that $X$ kings were removed, there are $4-X$ kings remaining in the deck of 26 cards.
We draw 3 cards from these 26 cards. The total number of ways to draw 3 cards is $\binom{26}{3}$.
The number of ways to draw 3 kings from the $4-X$ kings is $\binom{4-X}{3}$.
The probability of drawing 3 kings from the remaining 26 cards, given that $X$ kings were removed, is $P(\text{3 kings} | X) = \frac{\binom{4-X}{3}}{\binom{26}{3}}$.
Note that $\binom{4-X}{3} = 0$ if $4-X < 3$, i.e., if $X > 1$. So, we only need to consider the cases $X=0$ and $X=1$.

The total probability of drawing 3 kings is given by the law of total probability:
$P(\text{3 kings}) = \sum_{X=0}^{4} P(\text{3 kings} | X) P(X)$
$P(\text{3 kings}) = P(\text{3 kings} | X=0) P(X=0) + P(\text{3 kings} | X=1) P(X=1)$ (since $P(\text{3 kings} | X) = 0$ for $X \ge 2$)

For $X=0$:
$P(X=0) = \frac{\binom{4}{0} \binom{48}{26}}{\binom{52}{26}}$
$P(\text{3 kings} | X=0) = \frac{\binom{4}{3}}{\binom{26}{3}}$
Term for $X=0$: $\frac{\binom{4}{3}}{\binom{26}{3}} \frac{\binom{4}{0} \binom{48}{26}}{\binom{52}{26}} = \frac{4 \times 1 \times \binom{48}{26}}{\binom{26}{3} \binom{52}{26}}$

For $X=1$:
$P(X=1) = \frac{\binom{4}{1} \binom{48}{25}}{\binom{52}{26}}$
$P(\text{3 kings} | X=1) = \frac{\binom{3}{3}}{\binom{26}{3}}$
Term for $X=1$: $\frac{\binom{3}{3}}{\binom{26}{3}} \frac{\binom{4}{1} \binom{48}{25}}{\binom{52}{26}} = \frac{1 \times 4 \times \binom{48}{25}}{\binom{26}{3} \binom{52}{26}}$

$P(\text{3 kings}) = \frac{4 \binom{48}{26}}{\binom{26}{3} \binom{52}{26}} + \frac{4 \binom{48}{25}}{\binom{26}{3} \binom{52}{26}}$
$P(\text{3 kings}) = \frac{4}{\binom{26}{3} \binom{52}{26}} \left( \binom{48}{26} + \binom{48}{25} \right)$
Using the identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$, we have $\binom{48}{26} + \binom{48}{25} = \binom{49}{26}$.
$P(\text{3 kings}) = \frac{4 \binom{49}{26}}{\binom{26}{3} \binom{52}{26}}$

Now, let's expand the combinations:
$\binom{26}{3} = \frac{26 \times 25 \times 24}{3 \times 2 \times 1} = 26 \times 25 \times 4 = 2600$
$\binom{49}{26} = \frac{49!}{26! 23!}$
$\binom{52}{26} = \frac{52!}{26! 26!}$

$P(\text{3 kings}) = \frac{4 \times \frac{49!}{26! 23!}}{\frac{26 \times 25 \times 24}{6} \times \frac{52!}{26! 26!}}$
$P(\text{3 kings}) = \frac{4 \times 49! \times 6 \times 26! \times 26!}{26! 23! \times 26 \times 25 \times 24 \times 52!}$
$P(\text{3 kings}) = \frac{24 \times 49! \times 26!}{23! \times 26 \times 25 \times 24 \times 52!}$
Cancel out 24:
$P(\text{3 kings}) = \frac{49! \times 26!}{23! \times 26 \times 25 \times 52!}$
Expand factorials: $26! = 26 \times 25 \times 24 \times 23!$ and $52! = 52 \times 51 \times 50 \times 49!$
$P(\text{3 kings}) = \frac{49! \times (26 \times 25 \times 24 \times 23!)}{23! \times 26 \times 25 \times (52 \times 51 \times 50 \times 49!)}$
Cancel out $49!$, $23!$, $26$, $25$:
$P(\text{3 kings}) = \frac{24}{52 \times 51 \times 50}$
$P(\text{3 kings}) = \frac{24}{52 \times 51 \times 50} = \frac{12}{26 \times 51 \times 50} = \frac{6}{13 \times 51 \times 50} = \frac{3}{13 \times 51 \times 25}$
Since $51 = 3 \times 17$:
$P(\text{3 kings}) = \frac{3}{13 \times (3 \times 17) \times 25} = \frac{1}{13 \times 17 \times 25}$

$13 \times 17 = 221$
$13 \times 17 \times 25 = 221 \times 25 = 5525$.
The probability is $\frac{1}{5525}$.

Comparing this with the given options:
(A) $\frac{1}{(25)(17)(13)} = \frac{1}{25 \times 17 \times 13} = \frac{1}{5525}$
(B) $\frac{1}{(25)(15)(13)} = \frac{1}{4875}$
(C) $\frac{1}{(52)(17)(13)} = \frac{1}{11492}$
(D) $\frac{1}{(25)(51)(13)} = \frac{1}{16575}$

Option (A) matches our calculated probability.

The final answer is $\boxed{\frac{1}{(25)(17)(13)}}$.

Explanation of the solution:
Let $X$ be the number of kings removed among the first 26 cards. The number of remaining kings is $4-X$.
The probability of removing $X$ kings is $P(X) = \frac{\binom{4}{X}\binom{48}{26-X}}{\binom{52}{26}}$.
Given $X$, the probability of drawing 3 kings from the remaining 26 cards is $P(3K|X) = \frac{\binom{4-X}{3}}{\binom{26}{3}}$.
The total probability is $P(3K) = \sum_{X=0}^4 P(3K|X)P(X)$.
$P(3K) = P(3K|0)P(0) + P(3K|1)P(1)$ (since $P(3K|X)=0$ for $X \ge 2$).
$P(3K) = \frac{\binom{4}{3}}{\binom{26}{3}} \frac{\binom{4}{0}\binom{48}{26}}{\binom{52}{26}} + \frac{\binom{3}{3}}{\binom{26}{3}} \frac{\binom{4}{1}\binom{48}{25}}{\binom{52}{26}}$
$P(3K) = \frac{1}{\binom{26}{3}\binom{52}{26}} \left( \binom{4}{3}\binom{48}{26} + \binom{4}{1}\binom{48}{25} \right)$
$P(3K) = \frac{1}{\binom{26}{3}\binom{52}{26}} \left( 4\binom{48}{26} + 4\binom{48}{25} \right)$
$P(3K) = \frac{4}{\binom{26}{3}\binom{52}{26}} \left( \binom{48}{26} + \binom{48}{25} \right) = \frac{4\binom{49}{26}}{\binom{26}{3}\binom{52}{26}}$
Substitute the values of combinations and simplify:
$\binom{26}{3} = 2600$
$\binom{49}{26} = \frac{49!}{26!23!}$
$\binom{52}{26} = \frac{52!}{26!26!}$
$P(3K) = \frac{4 \times \frac{49!}{26!23!}}{\frac{2600}{1} \times \frac{52!}{26!26!}} = \frac{4 \times 49! \times 26! \times 26!}{26! 23! \times 2600 \times 52!}$
$P(3K) = \frac{4 \times 49! \times (26 \times 25 \times 24 \times 23!)}{23! \times 2600 \times (52 \times 51 \times 50 \times 49!)}$
$P(3K) = \frac{4 \times 26 \times 25 \times 24}{2600 \times 52 \times 51 \times 50} = \frac{4 \times 26 \times 25 \times 24}{(26 \times 100) \times 52 \times 51 \times 50}$
$P(3K) = \frac{4 \times 25 \times 24}{100 \times 52 \times 51 \times 50} = \frac{100 \times 24}{100 \times 52 \times 51 \times 50} = \frac{24}{52 \times 51 \times 50}$
$P(3K) = \frac{24}{52 \times 51 \times 50} = \frac{12}{26 \times 51 \times 50} = \frac{6}{13 \times 51 \times 50} = \frac{3}{13 \times 51 \times 25} = \frac{3}{13 \times (3 \times 17) \times 25} = \frac{1}{13 \times 17 \times 25}$.

The final answer is $\frac{1}{13 \times 17 \times 25}$.

The final answer is $\boxed{\frac{1}{(25)(17)(13)}}$.