Question

Question: What is the probability that a five-card poker hand contain the two of diamonds, three of spades, the six of hearts, the ten of clubs, and the king of hearts.

Short answer

Answer

The probability that a five-card poker hand contain the two of diamonds, three of spades, the six of hearts, the ten of clubs, and the king of hearts is$P\left(E\right)=\frac{1}{{}^{52}{C}_5}$

Step-by-step solution

 Given

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

The Concept of Probability

If represents 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

As per the problem we have been asked to find the probability that a five-card poker hand contain the two of diamonds, three of spades, the six of hearts, the ten of clubs, and the king of hearts.

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:

.

The probability that a randomly chosen card is one of these is$\frac{5}{52}$.

Now the rest is four to choose and fifty-one in total, so the probability that a randomly chosen card is one of these is$\frac{4}{51}$.

Now the rest is three to choose and fifty in total, so the probability that a randomly chosen card is one of these is$\frac{3}{50}$.

Now the rest is two to choose and forty-nine in total, so the probability that a randomly chosen card is one of these is$\frac{2}{49}$.

Now the rest is one to choose and forty-eight in total, so the probability that a randomly chosen card is one of these is$\frac{1}{48}$.

So, we have

$$\begin{gathered} P\left(E\right)=\frac{5}{52}\times \frac{4}{51}\times \frac{3}{50}\times \frac{2}{49}\times \frac{1}{48} \\ P\left(E\right)=\frac{5\times 4\times 3\times 2\times 1}{52\times 51\times 50\times 49\times 48} \\ P\left(E\right)=\frac{5\times 4\times 3\times 2\times 1}{52\times 51\times 50\times 49\times 48}\times \frac{47!}{47!} \\ P\left(E\right)=\frac{5!\times 47!}{52!} \end{gathered}$$

Solve Further we get,

$$\begin{gathered} P\left(E\right)=\frac{1}{\frac{52!}{5!\times 47!}} \\ P\left(E\right)=\frac{1}{52}{C}_5 \end{gathered}$$

The probability that a five-card poker hand contain the two of diamonds, three of spades, the six of hearts, the ten of clubs, and the king of hearts is$P\left(E\right)=\frac{1}{{}^{52}{C}_5}$