Question

recent research suggests that 47% of residents from a certain region have a home phone, 93% have a cell phone, and 41% of people have both. what is the probability that a resident from the region has
a) a home or cell phone?
b) neither a home phone nor a cell phone?
c) a cell phone but no home phone?
a) the probability of a resident having a home or cell phone is
(type an integer or decimal rounded to two decimal places as needed.)

Explanation:

Step1: Recall the formula for the probability of the union of two events

Let $A$ be the event of having a home - phone and $B$ be the event of having a cell - phone. The formula for $P(A\cup B)$ is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
Given $P(A) = 0.47$, $P(B)=0.93$, and $P(A\cap B)=0.41$.
$P(A\cup B)=0.47 + 0.93-0.41$

Step2: Calculate the value of $P(A\cup B)$

$P(A\cup B)=0.47+0.93 - 0.41=0.99$

Step3: Calculate the probability of having neither a home phone nor a cell phone

The probability of the complement of $A\cup B$ is $P(\overline{A\cup B})=1 - P(A\cup B)$.
Since $P(A\cup B)=0.99$, then $P(\overline{A\cup B})=1 - 0.99 = 0.01$

Step4: Calculate the probability of having a cell phone but no home phone

The probability of having a cell phone but no home phone is $P(B\cap\overline{A})=P(B)-P(A\cap B)$.
$P(B\cap\overline{A})=0.93 - 0.41=0.52$

Answer:

a) $0.99$
b) $0.01$
c) $0.52$