The Probability Calculus course exam has 3 questions and lasts 3 days. for a student be approved, you must correctly answer at least 2 questions. Each day has different versions of the questions and with different difficulties. The probability of correctly answering each of the three questions on the first day is $\frac{3}{4}$, on the second day it is $\frac{1}{2}$ and on the third day it is $\frac{1} {3}$. If the student will answer with probabilities $\frac{1}{6}$ the test on the first day, $\frac{2}{6}$ on the second and $\frac{3}{6}$ on the third, calculate the student's expectation of passing, if each question has the same value.
So, I found this question, the question is kind of confusing.
What I understood.
A student has 3 days of exams with 3 questions per day, ie 9 questions. The student needs to get 2 questions right out of 9 and he can ask a maximum of 3 questions and these questions can be answered in any order (3 on day 1, 1 on day 2 and 2 on day 3, etc). The probability that the student gets any of the questions right on the first day is $\frac{3}{4} = x$ on the second day it is $\frac{1}{2} = y$ and on the third day it is $ \frac{1}{3} = z$. We also have the probability that the student will answer the test on a given day, the probability of answering the test on the first day is $\frac{1}{6}$, on the second day it is $\frac{2}{6}$ and on the second day it is $\frac{2}{6}$. third is from $\frac{3}{6}$.
Assuming that (from what I understand from the question) the probability of answering each of the questions is $\frac{1}{6}= a$ on day 1, $\frac{2}{6} = b $ on day 2 and $\frac{3}{6} = c$ on day 3. Probability of getting the answers right is $\frac{3}{4}= x$ on day 1, $\frac{1}{2} = y $ on day 2 and $\frac{1}{3} = z $ on day 3.
case 1
All 3 questions are correct
$a*x*\binom{3}{1}*b*y*\binom{3}{1}*c*z*\binom{3}{1} + a*x*\binom{3}{2}*b*y*\binom{3}{1} + a*x*\binom{3}{2} *c*z*\binom{3}{1} + b*y*\binom{3}{2}*c*z*\binom{3}{1} + b*y*\binom{3}{1}*c*z*\binom{3}{2} + a*x*\binom{3}{3} + b*y*\binom{3}{3} + c*z*\binom{3}{3} +$
case 2
Only 2 questions right
$a*x*\binom{3}{1}*b*y*\binom{3}{1} + a*y*\binom{3}{1}*c*z*\binom{3}{1} + b*y*\binom{3}{1}*c*z*\binom{3}{1} + a*x*\binom{3}{2} + b*y*\binom{3}{2} + c*z*\binom{3}{2} $
$\frac{case 1 + case 2}{\binom{9}{2}+\binom{9}{3}}$
I think I didn't understand the question right and I have no confidence in my answer if anyone can help me, thanks in advance.
Let $P_1$ represent probability of passing on first day.
$P_1 = \frac{1}{6}$ (P(getting two q right) + P(getting 3 q right)
$= \frac{1}{6} \left(\binom{3}{2} \left(\frac{3}{4}\right)^2 \frac{1}{4} + \left(\frac{3}{4}\right)^3\right) = \frac{9}{64}$
Similarly for $P_2$ and $P_3$ and add them up.