A hamburger chain found that 75% of all customers use mustard, 80% use ketchup, and 65% use both, when ordering a hamburger. What are the probabilities that

a)A ketchup-user uses mustard?
b)A mustard-user uses ketchup?


Thirty percent of the managers in a certain company have MBA degrees as well as professional training. Eighty percent of all managers in the firm have professional training. If a manager is randomly chosen and found to have professional training, what is the probability that he or she also has an MBA?

Please do not double post questions. https://www.askmehelpdesk.com/mathematics/probability-583030.html

This type of problem uses what's called "conditional probability". You're trying to find the probability of some event happening, given that another event has happened. The formula is

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

where the left side reads as "The probability of A, given B". The numerator on the right side is the intersection of A and B (in other words, the probability of both events occurring).

So in this case, in part (a) you're asked to find the probability of a person using mustard, given that you already know the person uses ketchup. So for the above equation, using mustard is event A, and using ketchup is event B. The conditional probability is:

P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{0.65}{0.8}=81.25%.

Now can you do part (b)?

Your second question is very similar. Event A is that a person has an MBA. Event B is that they have professional training. You're being asked to find the conditional probability of A, given B.