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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

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 class=

Sagot :

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

[tex]\begin{gathered} n(\text{club) = 13} \\ n(\text{spade) =13} \\ n(\text{Heart) = 13} \\ n(Diamond)=\text{ 13} \\ \text{Total = 52} \end{gathered}[/tex]

Probability of an event is given as

[tex]Pr=\frac{Number\text{ of }desirable\text{ outcome}}{Number\text{ of total outcome}}[/tex]

Probability of choosing a club is evaluated as

[tex]\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}[/tex]

Probability of choosing a spade, without replacement

[tex]\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}[/tex]

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

[tex]\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}[/tex]

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

[tex]\frac{13}{204}[/tex]

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