Help with probability question I need help with the following:
A student takes a test of 10 questions.
he has a $0.6$ chance of answering correctly (independently from other questions).
Given that the student answered 8 questions correctly, what is the probability that he correctly answered 4 out of the first 5 questions?
this is what I've done so far:
Let $p=0.6$.
$A$ — the student correctly answered 4 questions out of the first 5.
$B$ — the student answered 8 questions correctly.
so I need to find $P(A \mid B)$. Obviously these two are dependent, so:
$$P(A \mid B)=\frac{P(A\cap B)}{P(B)}$$
now:
"$A$ intersection with $B$", is actually: "The student answered 8 questions correctly, and 4 of them were in the first 5", so:
$$P(A\cap B)=\binom{5}{4}p^4(1-p)\binom{5}{4}p^4(1-p)=\binom{5}{4}\binom{5}{4}p^8(1-p)^2$$
and:
$$P(B)=\binom{10}{8}P^8(1-p)^2$$
So I get:
$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{\binom{5}{4}\binom{5}{4}p^8(1-p)^2}{\binom{10}{8}P^8(1-p)^2}=\frac{\binom{5}{4}\binom{5}{4}}{\binom{10}{8}},$$
but that's obviously wrong... :(
I could really use some advices, tips and help here.
Thanks in advance.
 A: You identified correctly that the events are dependent, so we use $$P(A|B) = \frac{P(A \land B)}{P(B)}$$
To find $P(B)$, we use the Binomial Probability Formula: it gives $$P(B) = {10 \choose 8} \cdot 0.6^8 \cdot 0.4^2$$
To find $P(A\land B)$, we need to compute the probability that the student answered exactly four out of five questions correctly (first half of the test), and then answered exactly four out of the other five questions correctly (second half of the test). Note that by 'splitting up' the events like that, we can separately compute the probabilities and then just multiply them, as those two events are independent!
(Note that both events will have the same probability.) So by the Binomial Probability Formula, it'll be $$P(\text{Event}) = {5 \choose 4} \cdot 0.6^4 \cdot 0.4$$ So $$P(A \land B) = \left({5 \choose 4} \cdot 0.6^4 \cdot 0.4 \right)^2 = {5 \choose 4}^2 \cdot 0.6^8 \cdot 0.4^2$$
$$\therefore \frac{P(A\land B)}{P(B)} = \frac{{5 \choose 4}^2 \cdot 0.6^8 \cdot 0.4^2}{{10 \choose 8} \cdot 0.6^8 \cdot 0.4^2} = \frac{{5 \choose 4}^2}{{10 \choose 8}} = \frac59$$
Which is the same as the result you found. I think your answer is correct.
