Question

In an experiment, college students were given either four quarters or a $1 bill and they could either keep the money or spend it on gum. The results are summarized in the table. Complete parts (a) through (c) below. Purchased Gum Kept the Money Students Given Four Quarters 25 15 14 Students Given a $1 Bill 29 a. find the probability of randomly selecting a student who spent the money, given that the student was given a $1 bill. The probability is (Round to three decimal places as needed.) b. find the probability of randomly selecting a student who kept the money, given that the student was given a $1 bill. The probability is (Round to three decimal places as needed.) c. what do the preceding results suggest? A. A student given a $1 bill is more likely to have kept the money. B. A student given a $1 bill is more likely to have spent the money than a student given four quarters. C. A student given a $1 bill is more likely to have kept the money than a student given four quarters. D. A student given a $1 bill is more likely to have spent the money.

Answer 1

Answer:

a) $P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

b) $P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

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:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{40}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{15}{83}$

And if we replace we got:

$P(A|B) = \frac{15/83}{44/83} =\frac{15}{44}=0.341$

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:

$P(A|B) =\frac{P(A and B)}{P(B)}$

We have that $P(A)= \frac{43}{83} , P(B)= \frac{44}{83}, P(A and B)= \frac{29}{83}$

And if we replace we got:

$P(B|A) = \frac{29/83}{44/83} =\frac{29}{44}=0.659$

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