If she gets the first problem correct, what is the probability she gets an A? Problem statement -
Jane must get at least 3 of the 4 problems on the exam correct to get an A. She has been able to do 80% of the problems on old exams so she assumes that the probability she gets any correct is 0.8. She also assumes that the results on different problems is independent.
a) What is the probability she gets an A?
b) If she gets the first problem correct, what is the probability she gets an A?
My attempt -
a)
p = .8
(1-p) = .2
n = 4
k = 3, 4
A = {she gets an A}
P(A) = $4(.8)^3(.2) + (.8)^4 = .8192$
b) This is the part where I'm stuck at. I know I have to use Bayes' theorem, but I don't know how to find all of the relevant probabilities.
B = {first problem correct}
We want to find P(A | B). We know that
$P(A | B)$ = $\frac{P(B | A)P(A)}{P(B | A)P(A) + P(B | A^C)P(A^C)}$
But I don't know how to find $P(B | A)$ and $P(B | A^C)$.
How do I find $P(B | A)$ and $P(B | A^C)$?
 A: You have the first part correct.$$\mathsf P(A)=\binom 43p^3q+\binom 44p^4$$

Now, when given that she gets the first question correct (event $Q_1$), then she will get an A should she get at least 2 of the 3 remaining questions correct.  So the second verse is much the same as the first.
$$\mathsf P(A\mid Q_1)=\binom 32p^2q+\binom 33p^3$$
A: I don't think Bayes' theorem is going to be useful here. In fact, if I wanted to find $P(B|A)$, I'd probably start by finding $P(A|B)$ first, which is the reverse of what you've been trying to do.
However, you did a good job on the first part, and I think you'll be able to solve the second part with a hint.
Define $C_1$ as "she gets the first problem correct," $C_2$ as "she gets the second problem correct," and likewise for $C_3$ and $C_4$.

*

*What are $P(C_1|B)$, $P(C_2|B)$, $P(C_3|B)$, and $P(C_4|B)$?

*Are these four events independent?

*Now that you have the answers to the above two bullet points, do you have enough information to solve the problem?

