Mary knows the answers to $20$ of the $25$ multiple choice questions on the Psychology $101$ exam, but she has skipped several of the lectures, she must take random guesses for the other five. Assuming each question has four answers, what is the probability she will get exactly $3$ of the last $5$ questions right?
[Math] Probability problem with binomial/multinomial distribution
binomial-coefficientsprobability
Related Solutions
Let $U_k$ be an indicator random variable, equal to 1 if the $k$-th question has been guessed correctly. Clearly $(U_1, U_2,\ldots,U_n)$ are independent Bernoulli random variables with $\mathbb{E}\left(U_k\right) = \frac{1}{k+1}$.
The total number of correct guesses equals: $$ X = \sum_{k=1}^n U_k $$ The moment generating function of $X$ is easy to find: $$ \mathcal{P}_X\left(z\right) = \mathbb{E}\left(z^X\right) = \prod_{k=1}^n \mathbb{E}\left(z^{U_k}\right) = \prod_{k=1}^n \frac{k+z}{k+1} = \frac{1}{z} \frac{(z)_{n+1}}{(n+1)!} = \frac{1}{(n+1)!} \sum_{k=0}^n z^k s(n+1,k+1) $$ where $s(n,m)$ denote unsigned Stirling numbers of the first kind. Thus:$$ \mathbb{P}\left(X=m\right) = \frac{s(n+1,m+1)}{(n+1)!} [ 1 \leqslant m \leqslant n ] $$ The probability of getting at least $r$ equals: $$ \mathbb{P}\left(X \geqslant r\right) = \sum_{m=r}^{n} \frac{s(n+1,m+1)}{(n+1)!} $$
This reproduces your result for $n=4$ and $r=3$.
In[229]:= With[{n = 4, r = 3},
Sum[Abs[StirlingS1[n + 1, m + 1]]/(n + 1)!, {m, r, n}]]
Out[229]= 11/120
Here is the table for $1 \leqslant n \leqslant 11$:
[Math] On a question chosen at random, what is the probability that the student answers it correctly
For the first one, drawing a probability tree might help.
At first, you have the two branches of whether he knows the answer (0.7) and he doesn't know the answer (0.3).
If he knows the answer, then he answers correctly (1), so that, the probability that he knows and answer AND answers correctly is (0.7*1 =) 0.7.
If he doesn't know the answer, he has a probability of (1/5 =) 0.2 of getting the right answer, so that the probability that he doesn't know the answer AND answers correctly becomes: (0.3*0.2 =) 0.06
The rest should be a little easier.
Best Answer
Hint: Each question she guesses on is a trial. ... successes in ... independent trials with probability of success ... in each.