Question

A high school guidance counselor determined the following information regarding students and their likelihood of playing a sport and/or having a part-time job:30% of students play a sport. 65% of students have a part-time job. Of those students who play a sport, 50% have a job.What is the probability a person has a job, if they don't play a sport?

Answer 1

Conditional Probability

Given two events A and B (not excluding), the probability that A occurs given that B has occurred is called a conditional probability and is calculated as:

$P(A|B)=\frac{P(A\cap B)}{P(B)}$

Where P(A∩B) is the probability that A and B occur simultaneously and P(B) is the probability that B occurs.

Now with the given data, we must find the values of the required probabilities.

30% of the students play a sport (S), this means that:

70% of the students don't play a sport (NS).

65% of the students have a job.

Note that there could be students who both play sports and have a job.

Of the 30% of the students who play a sport, 50% have a job. This means that:

15% of the students play a sport and don't have a job

15% of the students play a sport AND have a job

65% - 15% = 50% of the students have a job and don't play a sport

That last number is the numerator of the equation given above:

P(A∩B) = 0.5

The event B corresponds to students that don't play a sport (NS), thus:

P(B) = 0.7

Thus we have:

$P(A|B)=\frac{0.5}{0.7}=\frac{5}{7}$

The required probability is 5/7 or 0.7143