Five cards are dealt from a shuffled deck. What is the probability that they are all of the same suit? That they are all diamond? That they are all face cards? That the five cards are a sequence in the same suit (for example, 3, 4, 5, 6, 7 of hearts)?

Five cards are drawn from a well shuffled deck of 52 cards.

When the order of arrangement is definite, the permutation is applied and when the order is not definite,combination is applied.

5 cards from the deck can be selected from 52 cards in $C\left(52,5\right)$ways. 5 cards from each suit can be selected in $C\left(13,5\right)$ways and a suit can be selected in $C\left(4,1\right)$.

Find the probability that all 5 cards are of the same suit

$$\begin{gathered} P\left(SameSuit\right)=\frac{C\left(4,1\right)\times C\left(13,5\right)}{C\left(52,5\right)} \\ =\frac{{\displaystyle \frac{4!}{31!}}\times {\displaystyle \frac{13!}{5!8!}}}{{\displaystyle \frac{52!}{47!5!}}} \\ =1.98\times {10}^3 \end{gathered}$$

The desired probability is $1.98\times {10}^{-3}$.

5 cards from the deck can be selected from 52 cards in $C\left(52,5\right)$ways. 5 cards from diamond can be selected in $C\left(13,5\right)$ways

Find the probability that all 5 cards are of diamond suit.

role="math" localid="1654858942591" $$\begin{gathered} P\left(Diamond\right)=\frac{C\left(13,5\right)}{C\left(52,5\right)} \\ =\frac{{\displaystyle \frac{13!}{5!8!}}}{{\displaystyle \frac{52!}{47!5!}}} \\ =4.95\times {10}^{-4} \end{gathered}$$

The desired probability is $4.95\times {10}^{-4}$.

5 cards from face cards can be selected in $C\left(12,5\right)$ways

Find the probability that all 5 cards are face cards

$$\begin{gathered} P\left(FaceCards\right)=\frac{C\left(12,5\right)}{C\left(52,5\right)} \\ =\frac{{\displaystyle \frac{12!}{5!7!}}}{{\displaystyle \frac{52!}{47!5!}}} \\ =3.05\times {10}^{-4} \end{gathered}$$

The desired probability is $3.05\times {10}^4$.

There are 8 sequence in each suit and a suit can be selected in $C\left(4,1\right)$.

Find the probability that all 5 cards are in sequence.

role="math" localid="1654859411068" $$\begin{gathered} \mathit{P}\left(CardsareinSequence\right)=\frac{C\left(4,1\right)\times 8}{C\left(52,5\right)} \\ =\frac{{\displaystyle \frac{4!}{1!3!}}\times 8}{{\displaystyle \frac{52!}{47!5!}}} \\ =1.23\times {10}^{-5} \end{gathered}$$

The desired probability is $1.23\times {10}^{-5}$.