Probability That Something Occured If Something Happens

Question

You go to grocery store 'A' 80% of time, and only go to grocery store 'B' 20%. Store 'A' is out of bread 10% of time and store 'B' is out of bread 40% of the time. Suppose he goes to a store and they are out of bread. What is the probability he went to store 'A'?

Answer 1

Bayes’ theorem makes it easy and mechanical, but you can reason it out from first principles. There are four possible combinations of events: he goes to $A$ and finds bread, he goes to $A$ and finds no bread, he goes to $B$ and finds bread, or he goes to $B$ and finds no bread. Make a $2\times 2$ table with a slot for each possibility, and fill in the percentage of the time each possibility occurs. In my table below the items in black are the labels and the things that we know immediately: he goes to $A$ $80$% of the time and to $B$ $20$% of the time. Then I filled in the blue items. When he goes to $A$, $10$% of the time he finds no bread; since he goes to $A$ $80$% of the time, that’s $10$% of $80$%, or $8$% of the time that he goes to $A$ and finds no bread. The other blue calculation is similar. Now I can combine those two results to find that on $16$% of his bread-buying trips he finds no bread; that’s the green figure. This is actually enough information to solve the problem, but for the sake of completeness let me finish filling in the table; you may find it useful for some other problem.

If $80$% of his trips are to $A$, and $8$% of his trips are to and unsuccessful, his successful trips to $A$ must make up $80-8=72$% of all of his bread-buying trips; that’s the brown entry in the upper lefthand corner of the table, and the brown entry immediately below it is calculated similarly. I can then get the total in the lower lefthand corner either by adding the two percentages above it or by subtracting the $16$% from $100$%; the fact that these numbers agree is an indication that I probably haven’t made any arithmetic errors.

$$\begin{array}{c|c|c|c} &\text{bread}&\text{no bread}&\text{Total}\\ \hline A&\color{brown}{80\%-8\%=72\%}&\color{blue}{10\%}\color{blue}{\text{ of }}\color{blue}{80\%=8\%}&80\%\\ \hline B&\color{brown}{20\%-8\%=12\%}&\color{blue}{40\%}\color{blue}{\text{ of }}\color{blue}{20\%=8\%}&20\%\\ \hline \text{Total}&\color{brown}{100\%-16\%=84\%\\72\%+12\%=84\%}&\color{green}{16}\%&100\% \end{array}$$

Now back to your problem. On $16$% of his bread-buying trips he finds no bread. $\frac8{16}=\frac12$ of these trips were to $A$, so the probability that a randomly chosen unsuccessful trip was to $A$ must be $\frac12$.