A survey found that 73% of adults have a landline at their residence (event A); 83% have a cell phone (event B). It is known that 2% of adults have neither a cell phone nor a landline. 3. What is the probability that an adult selected at random has both a landline and a cell phone? A. 0.58 B. 0.98 C. 0.6059 D. None of these Work: 4. Given an adult has a cell phone, what is the probability he does not have a landline?
A. 0.27
B. 0.25
C. 0.3012
D. None of these

In which a is the probability that an adult has a landline but not a cell phone and [tex]A \cap B[/tex] is the probability that an adult has both of these things.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

The sum of all the subsets is 1:

[tex]a + b + (A \cap B) + C = 1[/tex]

2% of adults have neither a cell phone nor a landline.

This means that [tex]C = 0.02[/tex].

73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)

So [tex]A = 0.73, B = 0.83[/tex].

What is the probability that an adult selected at random has both a landline and a cell phone?

This is [tex]A \cap B[/tex].

We have that [tex]A = 0.73[/tex]. So

[tex]A = a + (A \cap B)[/tex]

[tex]a = 0.73 - (A \cap B)[/tex]

By the same logic, we have that:

[tex]b = 0.83 - (A \cap B)[/tex].

So

[tex]a + b + (A \cap B) + C = 1[/tex]

[tex]0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1[/tex]

[tex](A \cap B) = 0.75 + 0.83 - 1 = 0.58[/tex]

So the answer for question 3 is A.

4. Given an adult has a cell phone, what is the probability he does not have a landline?

83% of the adults have a cellphone.

We have that

[tex]b = B - (A \cap B) = 0.83 - 0.58 = 0.25[/tex]

25% of those do not have a landline.

So [tex]P = \frac{0.25}{0.83} = 0.3012[/tex]

The answer for question 4 is C.