Suppose that a person becomes unemployed. What is the probability that this person also was college educated?

Question

I have this question that I am having a hard time understanding. And how I should go about regarding questions like this. Could someone explain how my thought process should be regarding questions like these or help me out?

Question:

Unemployment is affecting different groups differently. According to a American survey the unemployment on a national level was $13\%$, but only $6\%$ for people with a college education. Suppose that $21\%$ of the people in the workforce that is college educated. Now suppose that a person becomes unemployed. What is the probability that this person also was college educated?

The answer should be $9.7\%$, but I can't figure out how to get to this number.

Answer 1

Hints:

1. You need to calculate $$P(C|U) = \frac{P(C\cap U)}{P(U)}$$ where U, C are Unemployed and College grads respectively. (This is just the definition of conditional probability if you are not sure)
2. Now, we know 6% of C are in U. ie, $$|U\cap C| = 0.06|C|$$ And we know |C| is 21% of the population.
3. We also know |U| so you can calculate 1. to get ~9.69%

Answer 2

Let $U$ be the event that the person is unemployed.

Let $C$ be the event that the person is college educated.

We are given the following probabilities: \begin{align*} \Pr(U) & = \frac{13}{100}\\ \Pr(C) & = \frac{21}{100}\\ \Pr(U \mid C) & = \frac{6}{100} \end{align*} where $\Pr(U \mid C)$ is the probability that a person is unemployed given that she or he is college educated.

Then we are asked to calculate $\Pr(C \mid U)$, the probability that a person is college educated given that he or she is unemployed.

By definition, $$\Pr(C \mid U) = \frac{\Pr(C \cap U)}{\Pr(U)} \tag{1}$$ The probability that a person is both college educated and unemployed can be found by multiplying the probability that the person is college educated by the probability that he or she is unemployed given that he or she is college educated, that is, $$\Pr(C \cap U) = \Pr(C)\Pr(U \mid C)$$ Substituting $\Pr(C)\Pr(U \mid C)$ for $\Pr(C \cap U)$ in equation 1 yields $$\Pr(C \mid U) = \frac{\Pr(C)\Pr(U \mid C)}{\Pr(U)} \tag{2}$$ You can complete the problem by substituting the given numbers into equation 2.