Question

A survey of 249 people asks about their favorite flavor of ice cream. The results of this survey, broken down by the age group of the respondent and their favorite flavor, are as follows:
Chocolate Vanilla Strawberry
Children 40 10 44
Teens 34 10 38
Adults 17 43 13
If one person is chosen at random, find the probability that the person:______.
a) is an adult.
b) likes chocolate the best.
c) is an adult OR likes vanilla the best.
d) is a child AND likes vanilla the best.
e) likes strawberry the best, GIVEN that the person is a child.
f) is a child, GIVEN that the person likes strawberry the best.

Answer 1

Answer:

a) $P(Adult)=\frac{73}{249}=0.2932=29.32%$

b) $P(Chocolate)=\frac{91}{249}=0.3655=36.55%$

c) $P(AdultorVanilla)=\frac{31}{83}=0.3734=37.34%$

d) $P(ChildandVanilla)=\frac{10}{249}=0.0402=4.02%$

e) $P(Strawberry/Child)=\frac{22}{47}=0.4681=46.81%$

f) $P(Child/strawberry)=\frac{44}{95}=0.4632=46.32%$

Step-by-step explanation:

a)

In order to solve part a of the problem, we need to find the number of adults in the survey and divide them into the number of people in the survey by using the following formula>

$P=\frac{desired}{possible}$

In this case we have a total of 17+43+13 adults which gives us 73 adults and a total of 249 people surveyed so we get:

$P(Adults)=\frac{73}{249}=0.2932=29.32%$

b)

The same principle works for part b

there are: 40+34+17=91 people who likes chocolate ice cream the best so the probability is:

$P(Chocolate)=\frac{91}{249}=0.3655=36.55%$

c)

when it comes to the or statement, we can use the following formula:

P(A or B) = P(A) + P(B) - P( A and B)

In this case:

$P(Adult)=\frac{73}{249}$

$P(Vanilla)=\frac{10+10+43}{249}=\frac{63}{249}$

$P(AdultandVanilla)=\frac{43}{249}$

so:

$P(AdultorVanilla)=\frac{73}{249}+\frac{63}{249}-\frac{43}{249}$

$P(AdultorVanilla)=\frac{31}{83}=0.3734=37.34%$

d)

Is a child and likes vanilla the best.

In the table we can see that 10 children like vanilla so the probability is:

$P(ChildandVanilla)=\frac{10}{249}=0.0402=4.02%$

e)

Likes strawberry the best, GIVEN that the person is a child.

In this case we can make use of the following formula:

$P(B/A)=\frac{P(AandB)}{P(A)}$

so we can get the desired probabilities. First, for the probability of the person liking strawberry the best and the person being a child, we know that 44 children like strawberry the best, so the probability is:

$P(childrenandstrawberry)=\frac{44}{249}$

Then, we know there are 40+10+44=94 children, so the probability for the person being a child is:

$P(Child)=\frac{94}{249}$

Therefore:

$P(Strawberry/Child)=\frac{\frac{44}{249}}{\frac{94}{249}}$

$P(Strawberry/Child)=\frac{22}{47}=0.4681=46.81%$

f)

The same works for the probability of the person being a child given that the person likes strawberry the best.

First, for the probability of the person liking strawberry the best and the person being a child, we know that 44 children like strawberry the best, so the probability is:

$P(childrenandstrawberry)=\frac{44}{249}$

Then, we know there are 44+38+13 persons like strawberry, so the probability for the person liking strawberry is:

$P(Child)=\frac{95}{249}$

Therefore:

$P(Child/Strawberry)=\frac{\frac{44}{249}}{\frac{95}{249}}$

$P(Child/strawberry)=\frac{44}{95}=0.4632=46.32%$