# Probability Question?

I'm studying for my first stat exam, but I'm having trouble setting this problem up. I understand conditional probability whenever it is in a cross-tabulation table, such as P (B|E), but I don't know what numbers to use whenever I have a word problem. I've been going over the problems in my book to practice, but I'm stuck on this one:

Here's the question:

The U.S. Bureau of Labor Statistics publishes data on the benefits offered by small companies to their employees. Only 42% offer retirement plans while 61% offer life insurance. Suppose 33% offer both retirement plans and life insurance as benefits. If a small company is randomly selected, determine the following probabilities:

a. The company offers a retirement plan given that they offer life insurance.

I tried to set it up as a cross-tabulation table, but it didn't work.

• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Feb 25, 2016 at 19:33

when the question says 42% offer retirement plans it means that for 100 people the company offers retirement plans to 42 people so,

Event A -> Probability of a person getting retirement plan = (no of favourable cases)/(total no of cases) = 42/100 = 0.42

Event B ->Similarly you can get the probability for a person getting life insurance = 61/100 = 0.61

$$Event \hspace{1mm} A \cap B -> Similarly\hspace{1mm}we\hspace{1mm}can\hspace{1mm}get\hspace{1mm}the\hspace{1mm}probability\hspace{1mm}that\hspace{1mm}a\hspace{1mm}person\hspace{1mm}gets\hspace{1mm}both\hspace{1mm}life\hspace{1mm}insurance\hspace{1mm}and\hspace{1mm}life\hspace{1mm}benifits\hspace{1mm}=\hspace{1mm}33/100\hspace{1mm}=\hspace{1mm}0.33$$

Now you have $$P(A),\hspace{2mm} P(B)\hspace{2mm} and\hspace{2mm} P(A \cap B)\hspace{1mm} and\hspace{1mm}you\hspace{1mm}have\hspace{1mm}to \hspace{1mm}find\hspace{1mm} P(A/B)\hspace{1mm}which\hspace{1mm}can \hspace{1mm}be\hspace{1mm}done \hspace{1mm}by\hspace{1mm}Bayes\hspace{1mm}Rule.$$

Hope this helps.

I interpreted it as 42 companies out of 100 vs people. Using Bay's Rule $$P(Life and Retirement) = P( Retirement | Life ) * P( Life )$$ Solving $$P( Retirement | Life ) = \frac{P(Life and Retirement)}{P( Life )}$$ Plugging in values $$P( Retirement | Life ) = \frac{0.33}{0.61} \approx0.54$$