Question

two cards are drawn without replacement from a standard deck of 52 playing cards what is the probability of choosing a club and then without replacement a spade

Answer 1

occurringGiven a total of 52 playing cards, comprising of Club, Spade, Heart, and Spade.

$\begin{gathered} n(\text{club) = 13} \\ n(\text{spade) =13} \\ n(\text{Heart) = 13} \\ n(Diamond)=\text{ 13} \\ \text{Total = 52} \end{gathered}$

Probability of an event is given as

$Pr=\frac{Number\text{ of }desirable\text{ outcome}}{Number\text{ of total outcome}}$

Probability of choosing a club is evaluated as

$\begin{gathered} Pr(\text{club) = }\frac{Number\text{ of club cards}}{Total\text{ number of playing cards}} \\ Pr(\text{club)}=\frac{13}{52}=\frac{1}{4} \\ \Rightarrow Pr(\text{club) = }\frac{1}{4} \end{gathered}$

Probability of choosing a spade, without replacement

$\begin{gathered} Pr(\text{spade without replacement})\text{ = }\frac{Number\text{ of spade cards}}{Total\text{ number of playing cards - 1}} \\ =\frac{13}{51} \\ \Rightarrow Pr(\text{spade without replacement})=\frac{13}{51} \end{gathered}$

Thus, the probability of both events occuring (choosing a club, and then without replacement a spade) is given as

$\begin{gathered} Pr(\text{club) }\times\text{ }Pr(\text{spade without replacement}) \\ =\frac{1}{4}\text{ }\times\text{ }\frac{13}{51} \\ =\frac{13}{204} \end{gathered}$

Hence, the probability of choosing a club, and then without replacement a spade is

$\frac{13}{204}$