# Employment data at a large company reveal that 58 % of the workers are married, that 41 % are college graduates, and that 1/2 of the college graduates are married. What is the probability that...

Employment data at a large company reveal that 58 % of the
workers are married, that 41 % are college graduates, and
that 1/2 of the college graduates are married.
What is the probability that a randomly chosen worker is:
(b) Married but not a college graduate?
(c) Married or a college graduate?

embizze | Certified Educator

We are given that 58% of the workers are married and that 41% are graduates. Also, 1/2 of the graduates are married.

I. We can set up a probability tree:

Let G be the probability that a worker is a graduate, so G' is the probability that she is not a graduate. Let M be the probability that he is married and M' the probability of being single.

.41 G --------- .5 M       .205
|
-----------.5 M'      .205

.59 G' ---------    M
|
-----------     M'

Since the percentage of married workers is 58%, the sum of married graduates and married nongraduates is .58. Thus .41(.5)+.59x=.58 where x is the probability of being a married nongraduate. `x~~.6356 ` Then the probabilities in the right column of the tree sum to 1 (each employee is either a graduate or not, etc... So we have:

.41 G ------------ .5 M     .205
|
-------------- .5 M'    .205

.59 G' --------- .6356 M   .375
|
----------- .3644 M'   .215

(a) The probability of being married but not a graduate is (.59)(.6356)=.375

(b) The probability of being married or a graduate is 1 minus the complement or 1-(probability of being a nonmarried nongraduate) or 1-.215=.785

II. An alternative is to draw a Venn diagram -- two intersecting circles representing graduates and married employees.

For the graduates the circle total is .41; 1/2 are in the intersection so label the intersection .205 and G not in the intersection as .205. Since married is .58, and .205 is in the intersection, we have .58-.205=.375 in married but not in the intersection. This leaves .215 in the space outside either circle.

III. Use Bayesian probabilities:

`P(A|T)=(P(T|A)P(A))/(P(T|A)P(A)+P(T|A')(P(A'))) `

For (a) P(M|G')=`(P(G'|M)P(M))/(P(G'|M)P(M)+P(G'|M')P(M')) `