Question

recent research suggests that 45% of residents from a certain region have a home phone, 93% have a cell phone, and 43% 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 $P(A) = 0.45$, $P(B)=0.93$ and $P(A\cap B)=0.43$.
$P(A\cup B)=0.45 + 0.93-0.43$

Step2: Calculate the probability of having a home or cell phone

$P(A\cup B)=0.95$

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

The probability of the complement of $A\cup B$, denoted as $P((A\cup B)^C)$, is given by $P((A\cup B)^C)=1 - P(A\cup B)$. So $P((A\cup B)^C)=1 - 0.95=0.05$

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 $P(B\cap A^C)=P(B)-P(A\cap B)$. So $P(B\cap A^C)=0.93 - 0.43 = 0.50$

Answer:

a) $0.95$
b) $0.05$
c) $0.50$