Question

in a five - card poker hand, what is the probability of being dealt exactly one ace and no picture cards? click the icon to view a description of a standard deck of playing cards. the probability is (round to four decimal places as needed.)

Explanation:

Step1: Calculate total number of 5 - card hands

The total number of ways to choose 5 cards from a standard deck of 52 cards is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 52$ and $r=5$. So, $C(52,5)=\frac{52!}{5!(52 - 5)!}=\frac{52\times51\times50\times49\times48}{5\times4\times3\times2\times1}=2598960$.

Step2: Calculate number of ways to get exactly 1 ace and no picture - cards

There are 4 aces in a deck. The number of ways to choose 1 ace is $C(4,1)=\frac{4!}{1!(4 - 1)!}=4$.
There are 12 picture - cards (4 Jacks, 4 Queens, 4 Kings) in a deck. So the number of non - picture cards is $52-12 = 40$. We need to choose 4 non - picture cards from 40 non - picture cards. The number of ways to do this is $C(40,4)=\frac{40!}{4!(40 - 4)!}=\frac{40\times39\times38\times37}{4\times3\times2\times1}=91390$.
The number of 5 - card hands with exactly 1 ace and no picture cards is the product of the number of ways to choose 1 ace and the number of ways to choose 4 non - picture cards, so $C(4,1)\times C(40,4)=4\times91390 = 365560$.

Step3: Calculate probability

The probability $P$ of getting exactly 1 ace and no picture cards in a 5 - card hand is $P=\frac{\text{Number of favorable hands}}{\text{Total number of hands}}=\frac{365560}{2598960}\approx0.1406$.

Answer:

$0.1406$