# Conditional Probability Worksheet

Conditional Probability Worksheet
• Page 1
1.
The frequency table shows the results from a poll conducted about reading the newspaper in the morning.

Use the data to find P (Reads Newspaper | child)
 a. $\frac{15}{40}$ b. $\frac{17}{40}$ c. $\frac{23}{40}$ d. $\frac{12}{47}$

#### Solution:

The sample space for child is 17 + 23 = 40.

Among them, 17 read the newspaper in the morning.

So, P (Reads newspaper | Child) = 17 / 40

2.
The frequency table shows the results from a poll conducted about reading the newspaper in the morning.

Use the data to find P (Does not read Newspaper | Adult).
 a. $\frac{12}{17}$ b. $\frac{12}{87}$ c. $\frac{12}{40}$ d. $\frac{12}{47}$

#### Solution:

The sample space for adult is 35 + 12 = 47.

Among them, 12 do not read newspaper in the morning.

So, P (does not read newspaper | Adult) = 12 / 47

3.
The table shows the population information of the percentage of male and female population in the U.S., in the year 2000, according to a census.

 Under 18 Over 18 Male 26.8 73.2 Female 24.6 75.4

Using the table find:P (Male | Under 18).
 a. 2.2% b. 26.8% c. 18.03% d. 52.1%

#### Solution:

P (Male | under 18) = 26.8% / (26.8%+24.6%)

0.268(0.268 + 0.246) = 0.521

The probability that a person under 18 will be male is 52.1%.

4.
A math teacher gave her class two tests. 76% of the class passed the first test. 58% of the class passed both tests. Find the probability that a student who passed the second test, given that he passed the first test.
 a. 1.31 b. 0.58 c. 1 d. 0.76

#### Solution:

P (passes second test | passes first test) = P (Passes first and second test)P (Passes first test)
[Use P (B | A) = P (A and B)P (A).]

= 0.580.76 = 0.76

5.
The table shows the population information of the percentage of male and female population in the U.S., in the year 2000, according to a census.

Under 18

#### Solution:

P (Female | over 18) = 75.4%(73.2% + 75.4%) = 0.754(0.732 + 0.754)

= 0.507

The probability that a person over 18 is female is 50.7%.

6.
The frequency table shows the school enrolment (in thousands) in the year 2000. Let C represents california.
 Pre primary Elementary or high school College or graduate school California 1101 6472 2557 New Jersey 303 1444 470

Use the table to find P(C).
 a. 79% b. 82% c. 61% d. 91%

#### Solution:

The probability that a person enrolled in a school belongs to California = P(C)

P (C) = 1101 + 6472 + 25571101 + 6472 + 2557 + 303 + 1444 + 470

= 1013012347 = 0.8204

P (C) = 82.04% 82%

7.
The frequency table shows the school enrolment (in thousands) in the year 2000.
Let E represents elementary school.
 Pre Primary Elementary or high school College or graduate school California 1101 6472 2557 New Jersey 303 1444 470

Use the table to find P(E).
 a. 25.57% b. 11.01% c. 14.44% d. 64.11%

#### Solution:

P (E) = the probability that a person is enrolled in elementary school of California or New Jersey.

P (E) = 6472 + 14441101 + 6472 + 2557 + 303 + 1444 + 470 = 7916 / 12347 = 0.6411

P (E) = 64.11%

8.
The frequency table shows the school enrolment (in thousands) in the year 2000.
Let C represents california and G represents graduate school.
 Pre Primary Elementary or high school College or graduate school California 1101 6472 2557 New Jersey 303 1444 470

Use the table to find P(C and G).
 a. 20.7% b. 31.68% c. 25.24% d. 61.9%

#### Solution:

P (C and G) = 25571101 + 6472 + 2557 + 303 + 1444 + 470 = 2557 / 12347 = 0.207

The probability that a person enrolled in a graduate school of California is 20.7%

9.
The frequency table shows the school enrolment (in thousands) in the year 2000.
Let N represents New Jersey and R represents Pre primary school.
 Pre Primary Elementary or high school College or graduate school California 1101 6472 2557 New Jersey 303 1444 470

Use the table to find P(R | N).
 a. 76% b. 13.9% c. 18% d. 15.7%

#### Solution:

P(R and N) = 3031101 + 6472 + 2557 + 303 + 1444 + 470 = 303 / 12347 = 0.025

P(N) = 303 + 1444 + 4701101 + 6472 + 2557 + 303 + 1444 + 470 = 221712347 = 0.180

P(R | N) = 0.025 / 0.18 = 0.139
[Use P(R | N) = P(R and N)P(N).]

The probability that a person enrolls in primary school, given that the person is from New Jersey is 13.9%.

10.
The frequency table shows the school enrolment (in thousands) in the year 2000.
Let C represents California and G represents graduate school.
 Pre Primary Elementary or high school College or graduate school California 1101 6472 2557 New Jersey 303 1444 470

Use the table to find the probability that a person is from California, given that he is in graduate school.
 a. 64.7% b. 64.1% c. 84.5% d. 52.4%

#### Solution:

The required probability can be described by the notation P(C | G).

P (C | G) = P (C and G)P (G)
[Use P (B | A) = P (A and B)P (A).]

P (C and G) = 25571101 + 6472 + 2557 + 303 + 1444 + 470 = 255712347 = 0.207

P (G) = 2557 + 4701101 + 6472 + 2557 + 303 + 1444 + 470 = 302712347 = 0.245

P (C | G) = 0.2070.245 = 0.845

The probability that a person is from California given that he is in graduate school is 84.5%.