Probability question.
It is stated that a student is taking four modules. The probability that they pass any individual module depends on the fraction of lectures attended, f, such that the probability of passing a module is 0.95f.
a) If the student attends all lectures (i.e., f=1):
i) what is the probability they pass all four modules
ii) fail one module and pass the other three
iii) pass only one module
iv) pass no modules.
b) What is the probability that the student passes at least two modules?
c) Calculate the four probabilities from part a) on the assumption that the student attends only half the lectures.
d) What fraction of lectures must the student attend to have 40% chance of passing all four modules?
I don't have much experience in probability theory. What is the best approach to tackle this?
My attempt at part a)
I want to reason through by setting up some sort of probability. For example:
A= student passes module 1
B= passes module 2
C= passes module 3
D= passes module 4
So, I think it could be stated:
$P(A or B or C or D)= P(A)+P(B)+P(C)+P(D) - P(A&B) - P(A&C) - P(A&D) - P(B&C) - P(B&D) - (2*P(A&B&C&D))$
But there are more than four possible outcomes in total. For example: all 4 modules are passed, only one is passed, only two are passed, three are passed, none passed.
In that case, for part a), would one approach be to state:
i): $P(A+B+C+D) = \frac{4}{5}$
ii) $P(A+B+C) = \frac{3}{5}$
iii) $P(A) = \frac{1}{5}$
iv) $P(A) = \frac{0}{5}$
But I'm not sure what to do with this fraction, f, such that the probability of passing is 0.95*f*.
I wanted to try writing, for i), $P(A+B+C+D) = \frac{4}{5} = (1 - \frac{4}{5}) \cdot 0.95(1))$ but that doesn't seem to be correct.
Instead, I tried $P(A+B+C+D) = \frac{4}{5} = 0.95(1) \cdot \frac{4}{5})$. This seemed to return a more reasonable answer, but I am unsure if it is correct.
For parts b), c), d) I am also in need of guidance.
Any help guidance appreciated!
There are a few things that you should have picked up in the course or textbook you're following. The first is that this problem should be approached with a probability distribution called the binomial distribution. This distribution tells you the probability that exactly $k$ items are "chosen" out of a total of $n$ items, and is given by
$$ P(\text{$k$ items are chosen out of $n$}) = \binom{n}{k} p^k(1-p)^{n-k} $$
where $p$ is the probability that an individual item is chosen, and $\binom{n}{k}$ is a binomial coefficient and is given by
$$ \binom{n}{k} \stackrel{\text{def}}{=} \frac{n!}{k!(n-k)!} $$
This expression assumes (as we should in this problem, I presume) that the items are chosen independently of each other. If you're familiar with Pascal's triangle, $\binom{n}{k}$ is the $k+1$th entry in the $n+1$th row. For instance, if we consult the triangle below—
$$ \begin{array}{ccccccccccccc} & & & & & & 1 \\ & & & & & 1 & & 1 \\ & & & & 1 & & 2 & & 1 \\ & & & 1 & & 3 & & 3 & & 1 \\ & & 1 & & 4 & & 6 & & 4 & & 1 \\ & 1 & & 5 & & \boxed{10} & & 10 & & 5 & & 1 \\ 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 \end{array} $$
—we see that $\binom{5}{2} = \frac{5!}{2!3!} = 10$ is the third entry in the sixth row. Now, I put the word "chosen" in quotes, because it shouldn't be taken too literally: It really just means that some of the items satisfy some property, and the others don't. For instance, it could mean that some of the modules are passed by the student, and others aren't.
In this case, the probability that an individual module is passed is given by $p = 0.95f$, and initially at least, $f = 1$, so $p = 0.95$. Thus, for example, the probability that $k$ of the $n = 4$ modules are passed is given by the binomial distribution:
\begin{align} P(\text{$k$ modules are passed out of $n = 4$}) & = \binom{n}{k}p^k(1-p)^{n-k} \\ & = \binom{4}{k}(0.95)^k(0.05)^{4-k} \end{align}
If we wanted to know specifically the probability that the student passes two modules out of four, we set $k = 2$ and get
\begin{align} P(\text{$k = 2$ modules are passed out of $n = 4$}) & = \binom{4}{2}(0.95)^2(0.05)^2 \\ & = 6 \times 0.9025 \times 0.0025 \\ & = 0.0135375 \end{align}
Of course, I did this specific case because your problem asks for the other three. Part (i) asks for the case where $k = 4$, part (ii) asks for the case where $k = 3$, and I imagine you can figure out what $k$ equals for parts (iii) and (iv).
In part (b), you just need to add some of the subcases together.
In part (c), you just need to adjust $p$ because $f$ is no longer equal to $1$. (I imagine this is partly motivated by the object lesson of attending lectures?!)
In part (d), you need to first use the binomial distribution to express the probability of passing all four modules as a function of $p$. Then find out the minimum value of $p$ such that this probability is $0.4$. Finally, you use $p = 0.95f$ to find $f$ from $p$.