Question

a survey found that 57% of residents from a certain region have a home phone, 34% have a cell phone, and 16% of people have both. a) if a person from the region has a home phone, whats the probability that he or she has a cell phone also? b) are having a home phone and a cell phone independent events? explain c) are having a home phone and a cell phone mutually exclusive? explain a) the probability that a person from the region who has a home phone also has a cell phone is (round to three decimal places as needed.)

Explanation:

Step1: Define the probabilities

Let $H$ be the event of having a home - phone and $C$ be the event of having a cell - phone. We are given $P(H)=0.57$, $P(C)=0.34$, and $P(H\cap C)=0.16$.

Step2: Calculate the conditional probability for part a

The formula for conditional probability is $P(C|H)=\frac{P(H\cap C)}{P(H)}$. Substitute the given values: $P(C|H)=\frac{0.16}{0.57}\approx 0.281$.

Step3: Check for independence for part b

Two events $H$ and $C$ are independent if $P(H\cap C)=P(H)\times P(C)$. Calculate $P(H)\times P(C)=0.57\times0.34 = 0.1938$. Since $P(H\cap C)=0.16
eq0.1938 = P(H)\times P(C)$, having a home - phone and a cell - phone are not independent events.

Step4: Check for mutual exclusivity for part c

Two events $H$ and $C$ are mutually exclusive if $P(H\cap C) = 0$. Since $P(H\cap C)=0.16
eq0$, having a home - phone and a cell - phone are not mutually exclusive events.

Answer:

a) $0.281$
b) No. Because $P(H\cap C)=0.16$ and $P(H)\times P(C)=0.1938$, and $P(H\cap C)
eq P(H)\times P(C)$.
c) No. Because $P(H\cap C) = 0.16
eq0$.