Example

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.

Calculate the following probabilities using the table.

1. Find P(Person is a car phone user).
2. Find P(person had no violation in the last year).
3. Find P(Person had no violation in the last year AND was a car phone user).
4. Find P(Person is a car phone user OR person had no violation in the last year).
5. Find P(Person is a car phone user GIVEN person had a violation in the last year).
6. Find P(Person had no violation last year GIVEN person was not a car phone user)

Solution:

1. [latex]\displaystyle\frac{{\text{number of car phone users}}}{{\text{total number in study}}}=\frac{{305}}{{755}}\\[/latex]
2. [latex]\displaystyle\frac{{\text{number that had no violation}}}{{\text{total number in study}}}=\frac{{685}}{{755}}\\[/latex]
3. [latex]\displaystyle\frac{{280}}{{755}}\\[/latex]
4. [latex]\displaystyle{(\frac{{305}}{{755}}+\frac{{685}}{{755}})}-\frac{{280}}{{755}}=\frac{{710}}{{755}}\\[/latex]
5. [latex]\displaystyle\frac{{25}}{{70}}\\[/latex](The sample space is reduced to the number of persons who had a violation.)
6. [latex]\displaystyle\frac{{405}}{{450}}\\[/latex] (The sample space is reduced to the number of persons who were not car phone users.)

Example

This table shows a random sample of 100 hikers and the areas of hiking they prefer.

Hiking Area Preference

1. Complete the table.
2. Are the events “being female” and “preferring the coastline” independent events?Let F = being female and let C = preferring the coastline.
   1. Find P(F AND C).
   2. Find P(F)P(C)

   Are these two numbers the same? If they are, then
   F and C are independent. If they are not, then F and C are not independent.

3. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let
   M = being male, and let L = prefers hiking near lakes and streams.
   1. What word tells you this is a conditional?
   2. Fill in the blanks and calculate the probability: P(___|___) = ___.
   3. Is the sample space for this problem all 100 hikers? If not, what is it?
4. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks.
   1. Find P(F).
   2. Find P(P).
   3. Find P(F AND P).
   4. Find P(F OR P).

Solution:

1. Hiking Area Preference
   

   Sex	The Coastline	Near Lakes and Streams	On Mountain Peaks	Total
   Female	18	16	11	45
   Male	16	25	14	55
   Total	34	41	25	100

2. P(F AND C) = [latex]\displaystyle\frac{{18}}{{100}}[/latex] = 0.18
   P(F)P(C) = [latex]\displaystyle(\frac{{45}}{{100}})(\frac{{34}}{{100}})[/latex] = (0.45)(0.34) = 0.153
   P(F AND C) ≠ P(F)P(C), so the events F and C are not independent.
3. The word given tells you that this is a conditional. [latex]\displaystyle{P}{({M}|{L})}=\frac{{25}}{{41}}[/latex] No, the sample space for this problem is the 41 hikers who prefer lakes and streams.
4. 1. P(F) = [latex]\displaystyle\frac{{45}}{{100}}\\[/latex]
   2. P(P) = [latex]\displaystyle\frac{{25}}{{100}}\\[/latex]
   3. P(F AND P) = [latex]\displaystyle\frac{{11}}{{100}}\\[/latex]
   4. P(F OR P) = [latex]\displaystyle\frac{{45}}{{100}} + \frac{{25}}{{100}} - \frac{{11}}{{100}} = \frac{{59}}{{100}}\\[/latex]

Example

Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is [latex]\displaystyle\frac{{1}}{{5}}[/latex].

Door Choice

• The first entry [latex]\displaystyle\frac{{1}}{{15}}={(\frac{{1}}{{5}})}{(\frac{{1}}{{3}})}\\[/latex] is P(Door One AND Caught)
• The entry [latex]\displaystyle\frac{{4}}{{15}}={(\frac{{4}}{{5}})}{(\frac{{1}}{{3}})}\\[/latex] is P(Door One AND Not Caught)

Verify the remaining entries.

1. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
2. What is the probability that Alissa does not catch Muddy?
3. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

Solution:

1. Door Choice
   

   Caught or Not	Door One	Door Two	Door Three	Total
   Caught	[latex]\displaystyle\frac{{1}}{{15}}\\[/latex]	[latex]\displaystyle\frac{{1}}{{12}}\\[/latex]	[latex]\displaystyle\frac{{1}}{{6}}\\[/latex]	[latex]\displaystyle\frac{{19}}{{60}}[/latex]
   Not Caught	[latex]\displaystyle\frac{{4}}{{15}}\\[/latex]	[latex]\displaystyle\frac{{3}}{{12}}\\[/latex]	[latex]\displaystyle\frac{{1}}{{6}}\\[/latex]	[latex]\displaystyle\frac{{41}}{{60}}[/latex]
   Total	[latex]\displaystyle\frac{{5}}{{15}}\\[/latex]	[latex]\displaystyle\frac{{4}}{{12}}\\[/latex]	[latex]\displaystyle\frac{{2}}{{16}}\\[/latex]	1

2. [latex]\displaystyle\frac{{41}}{{60}}\\[/latex]
3. [latex]\displaystyle\frac{{9}}{{19}}\\[/latex]

example

This table contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.

United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

TOTAL each column and each row. Total data = 4,520.7

1. Find P(2009 AND Robbery).
2. Find P(2010 AND Burglary).
3. Find P(2010 OR Burglary).
4. Find P(2011|Rape).
5. Find P(Vehicle|2008).

Solution:

1. 0.0294
2. 0.1551
3. 0.7165
4. 0.2365
5. 0.2575