Respuesta :
Answer:
[tex]\text{a)} \frac{1}{17} \text{b)} \frac{1}{221}\text{c)}\frac{12}{13}\text{d)}\frac{188}{221}[/tex]
Step-by-step explanation:
GIVEN: You draw two cards from a standard [tex]52[/tex] card deck. You replace the first card and shuffle thoroughly before you draw the second.
TO FIND: a) What is the probability that the first card is [tex]7[/tex] b) What is the probability that both cards are
SOLUTION:
Let [tex]\text{A}[/tex] and [tex]\text{B}[/tex] be two events such that
[tex]\text{A}[/tex][tex]=\text{first card drawn is 7}[/tex]
[tex]\text{B}[/tex][tex]=\text{Second card is 7}[/tex]
[tex]\text{probability that first card is 7 =P(A)}[/tex]
[tex]\text{P(A)}=\frac{\text{total 7 numbered cards}}{\text{total cards in deck}}[/tex]
[tex]\text{P(A)}=[/tex] [tex]\frac{4}{52}=\frac{1}{13}[/tex]
a)
[tex]\text{probability that first card is 7 =P(A)}[/tex]
[tex]\text{P(A)}=\frac{1}{13}[/tex]
b)
[tex]\text{probability that both cards are 7}=[/tex][tex]\text{first card is 7}\times\text{second card is 7}[/tex]
[tex]\text{probability that both cards are 7}=[/tex][tex]\text{P(A).P(B)}[/tex]
[tex]\text{P(B)}=\frac{\text{total number of 7 numbered cards left}}{\text{total number of cards in deck left}}[/tex]
[tex]\text{P(B)}=\frac{3}{51}=\frac{1}{17}[/tex]
[tex]\text{probability that both cards are 7}=[/tex][tex]\frac{1}{13}\times\frac{1}{17}[/tex]
[tex]\text{probability that both cards are 7}=[/tex][tex]\frac{1}{221}[/tex]
c)
[tex]\text{probability that first card is not 7}=1-\text{probability that first card is 7}[/tex]
[tex]\text{probability that first card is not 7}=[/tex][tex]1-\text{P(A)}[/tex]
[tex]\text{probability that first card is not 7}=[/tex][tex]1-\frac{1}{13}[/tex]
[tex]\text{probability that first card is not 7}=[/tex][tex]\frac{12}{13}[/tex]
d)
[tex]\text{probability that both cards are not 7}=[/tex][tex](\text{first card is not 7}).(\text{second card is not 7})[/tex]
[tex]\text{probability that both cards are not 7}=[/tex][tex]\text{[1-P(A)].[1-P(B)]}[/tex]
[tex]\text{P(B)}=\frac{\text{total 7 numbered cards}}{\text{total cards left in deck}}[/tex]
[tex]\text{probability that both cards are not 7}=[/tex][tex](1-\frac{1}{13})\times(1-\frac{4}{51})[/tex]
[tex]\text{probability that both cards are not 7}=[/tex][tex]\frac{12}{13}\times\frac{47}{51}[/tex]
