Conditional probability on a multiple choice test A student is taking a multiple-choice test. Each question on the test has five possible answers and only one choice is correct. The probability this student knows the correct answer is 70%. If the student does not know the answer, they select an answer at random with each answer having an equal probability of being picked. Calculate the conditional probability that the student knew the answer given they answered the question correctly.
I started off like this:
Let B denote the event that the student knew the answer. Let A denote the event that the student answered the question correctly.
I was able to work out that $P(A)=\frac{3}{10}*\frac{1}{5} + \frac{7}{10} = \frac{19}{25}$
And I know the formula $P(B|A)=\frac{P(A \cap B)}{P(A)}$
I am unsure on how to work out $P(A \cap B)$. Any advice would be greatly apprectiated.
 A: It is easiest to draw a tree for these.

know        70%   -   correct   1       .7

                  /   correct   1/5     .06
don't know  30%   
                  \   incorrect 4/5     .24

Thus $P(A\cap B)=.7$ and your desired probability is $\frac{P(A\cap B)}{P(A)}=\frac{.7}{.76}= \frac{35}{38}$. It is very likely that the student knew the answer if they got it correct.
A: Let us denote the events: $A$: answered correctly, $B$: knew the answer.
We are provided that: $\mathsf P(B)=0.70$, $\mathsf P(A\mid B')=0.20$, and $\mathsf P(A\mid B)=1.00$.
NB: Thus ${\mathsf P(A\cap B)~{=\mathsf P(A\mid B)\,\mathsf P(B) \\= \mathsf P(B)\\=0.70}\\\mathsf P(A)\qquad~{=\mathsf P(B)+\mathsf P(A\mid B')\;\mathsf P(B')\\=0.70+0.20\cdot0.30\\=0.76}}$
We seek $\mathsf P(B\mid A)$.   Just use Bayes' Rule and the Law of Total Probability. $$\begin{align}\mathsf P(B\mid A) &=\dfrac{\mathsf P(A\mid B)\,\mathsf P(B)}{\mathsf P(A)} \\[1ex]&=\dfrac{\mathsf P(A\mid B)\,\mathsf P(B)}{\mathsf P(A\mid B)\;\mathsf P(B)+\mathsf P(A\mid B')\;\mathsf P(B')}\\&=\dfrac{35}{38}\end{align}$$
A: We are given that the student answered the question correctly, and we must find the probability that they actually knew the answer. Thus, we have two cases, one being that the student knew the answer and got it right, or the student did not know, and guessed correctly. The probability of the first case is $(7/10)$. The probability of the $2$nd case is $(3/10)(1/5)= 3/50$. Adding these up, we get that the total probability that the student answered the question correctly is $19/25$. Since we need to find the probability that the student KNEW the answer and got it correct, we do the probability he knew the answer and divide it by the total probability of getting the question right, so we get $(7/10)/(19/25)$ or $35/38$.
