What is the probability that a randomly selected person surveyed has a job, given that the person is less than 18 years old?
Solution: Let S be the Total number of people and S = 944
Let A be the number of people who have job and A = 607
[tex]\therefore P(A) = \frac{A}{S} = \frac{607}{944}[/tex]
Let B be the number of people who are less than 18 years old and B = 265
[tex]\therefore P(B) = \frac{B}{S} = \frac{265}{944}[/tex]
Also the number of people who have job and are less than 18 years old (A and B) = 20
[tex]\therefore P(A and B) = \frac{A and B}{S}=\frac{20}{944}[/tex]
We have to find the conditional probability of a person has a job given the person is less than 18 years old.
[tex]P(A|B) = \frac{P(A and B)}{P(B)}[/tex]
[tex]=\frac{\frac{20}{944} }{\frac{265}{944} }[/tex]
[tex]=\frac{20}{944} \times \frac{944}{265}[/tex]
[tex]=\frac{20}{265}=0.0755[/tex]
Therefore, the probability that a randomly selected person surveyed has a job, given that the person is less than 18 years old is 0.0755
