Question

Question: To Determine the probability that a five-card poker hand contain exactly one ace.

Short answer

Answer

The probability that a five-card poker hand contain exactly one ace is$P\left(E\right)=0.2995$

Step-by-step solution

 Given  

A pack of cards. In a pack of cards there are 52 cards.

The Concept of Probability 

Ifrepresents the sample space and$\mathit{E}$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:$\mathit{P}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

.

Determine the probability

Here, we have to find the probability that a five-card poker hand contain exactly one ace.

If $S$represents the sample space and $E$represents the event. Then the probability of occurrence of favourable event is given by the formula as below:

.$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

As we know that there are four Aces and a total of fifty-two cards.

We are supposed to draw five cards out of fifty-two cards but one ace out of four aces and four other cards out of remaining forty-eight and this can be accomplished as

$$\begin{gathered} P\left(E\right)=\frac{{}^4{C}_1{\times }^{48}{C}_4}{{}^{52}{C}_5} \\ P\left(E\right)=\frac{\frac{4!}{1!3!}\times \frac{48!}{4!44!}}{\frac{55!}{5!47!}} \\ P\left(E\right)=\frac{4\times 194,580}{2,598,960} \\ P\left(E\right)=0.2995 \end{gathered}$$

The probability that a five-card poker hand contain exactly one ace is $P\left(E\right)=0.2995$.