I am having trouble with a problem involving conditional probability. The question is as follows:
A student is taking a history quiz, which consists of $6$ multiple choice questions and each multiple choice question has $5$ answer options (and only $1$ of these $5$ answer options are correct). Suppose that the student learned the material asked in question $n$ with probability $p_n$; where $n =1,2, \ldots,6$. If the student learned the material asked in the question, then the student is guaranteed to answer that question. If they did not learn the material in question $n$, then that student guesses the answer to that question at random.
[a] What is the probability that the student learned all of the material covered on the exam, if they got a perfect score.
[b] What is the probability that the student got a perfect score 6/6 and did not learn any material that was covered on the exam?
[c] If $p_n = 1/2$, for all $n =1,2, \ldots,6$, find and simply the answers found in [a] and [b] as much as possible.
When beginning the problem, I first defined the following:
A: student answered $n$-th question correctly
B: student learned the material for $n$-th question
\begin{align*}
P[A|B]& =1;\\
P[A|B^C]& =1/5;
\end{align*}
where student chooses randomly.
$$P[B]=P_n;$$ where $n = 1,2, \ldots,6$
For part [a], I have the following:
P[student learned the material | answered correctly]
\begin{align*}
P[B|A] & = \frac{P[A|B]P[B]}{P[A|B]P[B]+P[A|B^C]P[B^C]}\\
& = \frac{P_n}{P_n +(1/5)(1-P_n)}
\end{align*}
Now, we assume that material for each question are different. So, each question is independent.
P[Student learned all material | answered all questions correctly]
$$\Pi_{n=1}^6 P[B|A]$$
$$\Pi_{n=1}^6 \frac{(P_n)}{P_n +(1/5)(1-P_n)}$$
For part [b], I have the following:
P[Student did not learn the material for nth question | they answered correctly]
$$=P[B^C\cap A]=P[A|B^C]P[B^C]$$
$$=P[B^C\cap A]=\frac{1}{5}(1-P_n)$$
P[Student did not learn any material | answered question correctly]
$$\Pi_{n=1}^6 \frac{1}{5}(1-P_n)$$
For part [c], I have the following:
[a] Becomes:
\begin{align*}
\Pi_{n=1}^6 \frac{.5}{.5+.2(1-.5)} & = \left(\frac{.5}{.5+.10}\right)^6\\
& =0.3349
\end{align*}
$$33.49\%$$
[b] Becomes:
$$\Pi_{n=1}^6 P_n = (\frac{1}{5}(1-P_n))^{6}=0.000001=0.0001\%$$
Any help or suggestions for this question and my answers would be greatly appreciated.
Best Answer
You misunderstood the question in part b). This is not a conditional probability like in part a). The “if” and “and” are bold for a reason; part a) asks for a conditional probability whereas part b) asks for the probability of the intersection of two events.
(Also, you seem to have made an error in copying the result for $P[B\mid A]$ from part a) to part b), but that’s not relevant to the actual question in part b).