Respuesta :
Answer:
(a)[tex]\frac{1}{17}[/tex] (b) [tex]\frac{1}{16}[/tex]
Step-by-step explanation:
GIVEN: Suppose that two cards are randomly selected from a standard [tex]52[/tex] card deck.
TO FIND: (a) What is the probability that the first card is a club and the second card is a club if the sampling is done without replacement? (b) What is the probability that the first card is a club and the second card is a club if the sampling is done with replacement.
SOLUTION:
(a)
Probability that first card is club [tex]P(A)=\frac{\text{total club cards}}{\text{total cards}}[/tex]
[tex]=\frac{13}{52}[/tex]
[tex]=\frac{1}{4}[/tex]
As sampling is done without replacement.
probability that second card is club [tex]P(B)=\frac{\text{total club cards}}{\text{total cards}}[/tex]
[tex]=\frac{12}{51}[/tex]
[tex]=\frac{4}{17}[/tex]
Probability that first card is club and second card is club [tex]=P(A)\times P(B)[/tex]
[tex]=\frac{1}{4}\times\frac{4}{17}=\frac{1}{17}[/tex]
(b)
Probability that first card is club [tex]P(A)=\frac{\text{total club cards}}{\text{total cards}}[/tex]
[tex]=\frac{13}{52}[/tex]
[tex]=\frac{1}{4}[/tex]
As sampling is done with replacement.
probability that second card is club [tex]P(B)=\frac{\text{total club cards}}{\text{total cards}}[/tex]
[tex]=\frac{13}{52}[/tex]
[tex]=\frac{1}{4}[/tex]
Probability that first card is club and second card is club [tex]=P(A)\times P(B)[/tex]
[tex]=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}[/tex]
