At the Campus Coffee Nook, 55% of the customers order regular coffee and 45% order flavored coffee. Of those who order regular coffee, 55% drink it black and 45% use sugar or cream. Of those who drink flavored coffee, 15% drink it black and 85% use sugar or cream.
I know that P(black coffee | regular)= .55
I know that P(black coffee and regular) = .303
How can I figure out P(black coffee) and P(flavored coffee | cream or sugar).
This solution uses Bayes' Theorem.
Let $X$ be the event that a customer orders flavored coffee. Then $\neg X$ is the event that a customer orders regular coffee.
Let $Y$ be the event that a customer orders coffee with cream and/or sugar. Then $\neg Y$ is the event that a customer orders black coffee.
We are given the following:
$$P(X) = 45\%$$ $$P(\neg X) = 55\%$$ $$P(\neg Y\text{ }|\text{ }\neg X) = 55\%$$ $$P(Y\text{ }|\text{ }\neg X) = 45\%$$ $$P(\neg Y\text{ }|\text{ }X) = 15\%$$ $$P(Y\text{ }|\text{ }X) = 85\%$$
The probability that a customer orders black coffee is $P(\neg Y)$. We know $$P(\neg Y) = P(X) \cdot P(\neg Y\text{ }|\text{ }X) + P(\neg X) \cdot P(\neg Y\text{ }|\text{ }\neg X)$$
We plug our known values in to this equation to get
$$P(\neg Y) = (0.45)(0.15)+(0.55)(0.55) = 0.37$$
We are asked for the probability that a customer orders flavored coffee, given that he/she orders it with cream and/or sugar. We want to find $P(X\text{ }|\text{ }Y)$.
First we find $P(Y)$. We know that $$P(Y) = P(Y\text{ }|\text{ }X) \cdot P(X) + (Y\text{ }|\text{ }\neg X) \cdot P(\neg X)$$
We plug in to solve for $P(Y)$:
$$P(Y) = (0.85)(0.45) + (0.45)(0.55) = 0.63$$
Bayes' Theorem states that $$P(X\text{ }|\text{ }Y) = \frac{P(Y\text{ }|\text{ }X) \cdot P(X)}{P(Y)}$$
Now we can just plug in the known values to get $$P(X\text{ }|\text{ }Y) = \frac{(0.85)(0.45)}{0.63} = \frac{17}{28} \approx 60.7\%$$
I noticed that you've asked several questions related to conditional probability that can all be solved using similar methods. I hope you're learning how to do problems like these in general $\ddot\smile$