Worded Conditional Probability

Question

A student has studied only 20% of the material for a module that is examined via multiple choice questions.

Given that a question covers the material he has studied, the student will definitely answer correctly.

Given that a question covers material that he has not studied, he picks an answer at random.

You can assume that a question either covers material he has studied, or not.

We give the student one multiple choice question with four choices and only 1 right answer.

Q1) given that the student answered correctly, what is the probability he picked the answer at random

My approach:

Let A=picked answer at Random

B=picked answer correctly

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

$P\left ( B \right )$=1/4

however i am unable to find out the numerator as I cant figure out $P(A)$

Answer 1

$P$(Correct Answer | Picked the answer at random) = $1/4$
$P$(Correct Answer | Picked the answer intelligently) = $1$

There are only two possibilities: Either the student Picked the answer at random, or intelligently. They can pick the answer intelligently only if it comes from that 20% syllabus that they studied. Assuming that the probability of question coming from the syllabus is uniform over the syllabus, the probability that they picked the answer intelligently is 20%, i.e. $20/100=1/5$
$P$(Picked the answer at random) = $4/5$
$P$(Picked the answer intelligently) = $1/5$

$P$(Correct Answer) = $(1/4\times4/5) + (1\times1/5)$ = $2/5$

P(Picked the answer at random | Correct Answer)
= $\dfrac{P(\text{Correct Answer | Picked the answer at random}) * P(\text{Picked the answer at random}) }{ P(\text{Correct Answer}) }$
= $\dfrac{(1/4 \times 4/5)}{(2/5)}$ = $1/2$

Answer 2

Let $A$ denote the event that the student got the question right because it was on material that he studied.

Let $B$ denote the event that the student got the question right.

Per Bayes Theorem, you have that

$$p(A|B) \times p(B) = p(A,B) \implies p(A|B) = \frac{p(A,B)}{p(B)}$$

$$= \frac{(.2) \times (1.0)}{[ ~(.2) \times (1.0) ~] ~+~ [ ~(0.8) \times (0.25) ~]}$$

$$= \frac{.2}{.2 + .2} = \frac{1}{2}.$$

Notice that your computation of $p(B)$ is wrong, and should in fact be $(0.4).$