Question

homework 34: problem 9 (1 point) a standard deck of 52 playing cards contains 4 aces and 4 kings. a person removes two aces and a king from a deck of 52 playing cards, so the deck only has 49 cards. then the person draws, without replacement, two more cards from the deck. find the probability that the person will draw two aces, two kings, or an ace and a king (in either order). (a standard 52 - card deck contains 4 aces and 4 kings.) answer:

Explanation:

Step1: Calculate total number of ways to draw 2 - card hands from 49 cards

The number of combinations of choosing 2 cards from 49 cards is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 49$ and $r=2$. So $C(49,2)=\frac{49!}{2!(49 - 2)!}=\frac{49\times48}{2\times1}=1176$.

Step2: Calculate number of ways to draw 2 aces

There are $4 - 2=2$ aces left in the deck. The number of ways to choose 2 aces from 2 aces is $C(2,2)=\frac{2!}{2!(2 - 2)!}=1$.

Step3: Calculate number of ways to draw 2 kings

There are $4 - 1 = 3$ kings left in the deck. The number of ways to choose 2 kings from 3 kings is $C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3\times2!}{2!×1!}=3$.

Step4: Calculate number of ways to draw 1 ace and 1 king

The number of ways to choose 1 ace from 2 aces and 1 king from 3 kings is $C(2,1)\times C(3,1)=\frac{2!}{1!(2 - 1)!}\times\frac{3!}{1!(3 - 1)!}=2\times3 = 6$.

Step5: Calculate favorable number of outcomes

The total number of favorable outcomes is the sum of the number of ways to draw 2 aces, 2 kings, and 1 ace and 1 king. So $1 + 3+6=10$.

Step6: Calculate the probability

The probability $P$ is the number of favorable outcomes divided by the number of total outcomes. So $P=\frac{10}{1176}=\frac{5}{588}$.

Answer:

$\frac{5}{588}$