Question

according to the u s. census bureau, 8.01% of 16 - to 24 - year - olds are high school dropouts. in addition, 2.06% of 16 - to 24 - year - olds are high school dropouts and unemployed. what is the probability that a randomly selected 16 - to 24 - year - old is unemployed, given they are a dropout?
the probability that a randomly selected 16 - to 24 - year - old is unemployed, given they are a dropout, is
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for conditional probability

The formula for conditional probability is \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). Let \(A\) be the event that a 16 - to 24 - year - old is a high - school dropout and \(B\) be the event that a 16 - to 24 - year - old is unemployed. Then \(P(A) = 0.0801\) (probability of being a dropout) and \(P(A\cap B)=0.0206\) (probability of being a dropout and unemployed).

Step2: Substitute the values into the formula

Substitute \(P(A) = 0.0801\) and \(P(A\cap B)=0.0206\) into the formula \(P(B|A)=\frac{P(A\cap B)}{P(A)}\). So \(P(B|A)=\frac{0.0206}{0.0801}\).

Step3: Calculate the value

\(\frac{0.0206}{0.0801}\approx0.257\)

Answer:

\(0.257\)