Probability question with two groups of students

Question

Thirty percent of the students in a calculus course and 20 percent of students in a statistics course receive A's. Furthermore, 60 percent of the students with an A in calculus receive an A in the statistics course. John received an A in the statistics course. Calculate the probability that he also received an A in the calculus course.

well since we know that out of 30% calc A's 18% are in the Stat A's group would it not follow that john has 18% chance of having an A in calc, or am i wrong?

Answer 1

Should be a straightforward application of Bayes' Theorem:

$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$

The question is asking: Given John got an A in stats, what is the probability he also received an A in calculus?

Let $A =$ an A in calculus
Let $B =$ an A in stats

We are given:
$P(A)=.3$
$P(B)=.2$
$P(B|A)=.6$

$$P(A|B)=\frac{(.6)(.3)}{.2}=.90$$

Or, $90\%$.

Answer 2

Let E be the event that a student receives an A in calculus, and

let F be the event that a student receives an A in statistics.

Then $P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{P(F|E)P(E)}{P(F)}=\frac{(.60)(.30)}{.20}=\frac{9}{10}$.

Answer 3

No, it does not follow that John has an 18% chance of getting an A in calculus.

Probability problems like this are especially difficult to reason through without defining variables and manipulating them symbolically.

Let $C_A = \{\text{the event that the student receives an A in Calculus}\}$, $S_A = \{\text{ the event that the student receives an A in Statistics}\}$ From these definitions and the problem statement we have $P(C_A) = 0.3$ and $P(S_A) = 0.2$. Additionally, we know that if a student received an A in Calculus, then the probability the same student received an A in the Statistics course is $0.6$ which gives us $P(S_A \vert C_A) = 0.6$

Now that we have appropriate variables with which to think about the problem let's go back and examine what is being asked. We are told John received an A in Statistics and want to know the probability he received an A in Calculus, that is we wish to know $P(C_A \vert S_A)$ which we can obtain without additional work from the given information by a straightforward application of Bayes Theorem.

$P(C_A\vert S_A) = \frac{P(S_A\vert C_A)P(C_A)}{P(S_A)}=\frac{0.6*0.3}{0.2} = 0.9$