Respuesta :
Answer:
a) [tex] P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341[/tex]
b) [tex] P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659[/tex]
c) A. A student given a $1 bill is more likely to have kept the money.
Because the probability 0.659 is atmoslt two times greater than 0.341
Step-by-step explanation:
Assuming the following table:
Purchased Gum Kept the Money Total
Students Given 4 Quarters 25 14 39
Students Given $1 Bill 15 29 44
Total 40 43 83
a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill.
For this case let's define the following events
B= "student was given $1 Bill"
A="The student spent the money"
For this case we want this conditional probability:
[tex] P(A|B) =\frac{P(A and B)}{P(B)}[/tex]
We have that [tex] P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}[/tex]
And if we replace we got:
[tex] P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341[/tex]
b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill.
For this case let's define the following events
B= "student was given $1 Bill"
A="The student kept the money"
For this case we want this conditional probability:
[tex] P(A|B) =\frac{P(A and B)}{P(B)}[/tex]
We have that [tex] P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}[/tex]
And if we replace we got:
[tex] P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659[/tex]
c. what do the preceding results suggest?
For this case the best solution is:
A. A student given a $1 bill is more likely to have kept the money.
Because the probability 0.659 is atmoslt two times greater than 0.341
