# If a student who received an A in probability is chosen at random, what is the probability that he/she also received an A in calculus?

This question has been asked before but the solution given was incorrect.(see here)

A prerequisite for students to take a probability class is to pass calculus. A study of correlation of grades for students taking calculus and probability was conducted. The study shows that 25% of all calculus students get an A, and that students who had an A in calculus are 50% more likely to get an A in probability as those who had a lower grade in calculus. If a student who received an A in probability is chosen at random, what is the probability that he/she also received an A in calculus?

Now here is my attempt: $C=$the event of an A in calculus and let $A=$ the event of an A in probability. Now we are given that $P(C)=.25$ and $P(A|C)=.5P(A|C^c)$ we want to find $P(C|A)$. We use the following formula $$P(C|A)=\frac{P(A|C)P(C)}{P(A)}$$ now the probability of $A$ is the following $$P(A)=P(A|C)P(C)+P(A|C^c)P(C^c)$$ when we plug this in we get the following $$P(C|A)=\frac{P(A|C)(.25)}{0.25(P(A|C)+2P(A|C)(.75)}=\frac{1}{7}.$$ Now the answer key in the book says the solution is $\frac{1}{3}$.

• $P(A|C)=1.5 P(A|C^c)$ – Qwerty Jun 23 '16 at 6:19
• Indeed. it was stated $50\%$ more likely. – Graham Kemp Jun 23 '16 at 6:55
• Why the duplicate? Asking for explanations on the other page was not enough? – Did Jun 23 '16 at 7:17
• @Did other solution is not correct $$P(A|B) \not = \frac{P(B)P(A|B)}{P(A)}$$ which they claim. – Andrew Jun 23 '16 at 16:33
• Yeah, an obvious typo. Anyway... Asking for explanations on the other page was not enough? – Did Jun 23 '16 at 16:49

Let's see what we have:

$P(A_{calc})=\frac{1}{4}$

$P(A_{prob}|A_{calc})=\frac{3}{2}p$, where $p=P(A_{prob}|A_{calc}^C)$ (complementary probability).

We can calculate this $p$, since $P(A_{prob}|A_{calc})+P(A_{prob}|A_{calc}^C)=1$. We have $1.5p+p=1\Rightarrow p=\frac{2}{5}$.

Now, we need to calculate $P(A_{calc}|A_{prob})$. Using Bayes' theorem, we have $$P(A_{calc}|A_{prob})=P(A_{prob}|A_{calc})\frac{P(A_{calc})}{P(A_{prob})}$$ $$\frac{3}{2}\cdot\frac{2}{5}\cdot \frac{1/4}{P(A_{prob}|A_{calc})P(A_{calc})+P(A_{prob}|A_{calc}^C)P(A_{calc}^C)}$$ Plugging everything in, we have $$\frac{3}{2}\cdot\frac{2}{5}\cdot \frac{1/4}{\frac{3}{5}\cdot\frac{1}{4}+\frac{2}{5}\cdot\frac{3}{4}}=\frac{1}{3}$$

• So 50 percent more likely is 1.5 more likely – Fernando Martinez Jul 17 at 21:00

The following is a more intuitive explanation of the problem that will not require the explicit use of Bayes formula.

The "number" (probability that student got an A) of students who got an A in calc is .25. The number that did not get an A in calc is .75.

We also know that the P(A in Stat | A in Calc) = 1.5P(A in Stat|~A in Calc).

We know that P(A Stat | A in calc) + P(A stat | ~A in calc) = 1.

By substitution, P(A Stat | ~A in Calc) + 1.5 P(A Stat | ~A Calc)= 1.

We get 5/2 P(A Stat| ~ A in Calc)= 1

So P(A Stat | ~A in Calc) = 2/5 and P(A Stat | A Calc)= 3/5

This tells us that the number of students who got an A in Calc and an A in Stats is 1/4 * 3/5 = 3/20. And the number of students who got an A in stats, but not an A in calc is 3/4 * 2/5 = 6/20.

This part that I just did can be easily understood by thinking about a tree diagram: Click the link to see the diagram

So there are 3/20 + 6/20 = 9/20 total As in Stat.

Out of the 9/20 As in stat, we already saw that 3/20 of them also had As in calc.

So we get that the P(A calc | A stat) = (3/20)/(9/20) = 3/9 = 1/3