Question

suppose there is a 22.3% probability that a randomly selected person aged 20 years or older is a smoker. in addition, there is a 11.2% probability that a randomly selected person aged 20 years or older is female, given that the person smokes. what is the probability that a randomly selected person aged 20 years or older is female and smokes? would it be unusual to randomly select a person aged 20 years or older who is female and smokes? the probability that a randomly selected person aged 20 years or older is female and smokes is (round to three decimal places as needed.)

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$, where $P(A|B)$ is the probability of event $A$ given event $B$, $P(A\cap B)$ is the probability of both $A$ and $B$ occurring, and $P(B)$ is the probability of event $B$ occurring. Let $A$ be the event that a person is female and $B$ be the event that a person is a smoker. We are given $P(B) = 0.223$ and $P(A|B)=0.112$.

Step2: Solve for $P(A\cap B)$

We can re - arrange the conditional - probability formula to solve for $P(A\cap B)$. Multiply both sides of the formula $P(A|B)=\frac{P(A\cap B)}{P(B)}$ by $P(B)$ to get $P(A\cap B)=P(A|B)\times P(B)$.
Substitute the given values: $P(A\cap B)=0.112\times0.223$.
$P(A\cap B)=0.112\times0.223 = 0.025$.

Answer:

$0.025$