Kdtyjb

A class of 30 students has 5 classes today, with J. being one of them.

• What is the chance, that J. will be the chosen student to explain the
  homework in at least two classes?


• What is the chance that someone will be chosen at least twice?

My solution is the following:
The probability for J. being chosen at least twice: $frac{1}{30}^2 cdot frac{29}{30}^3+frac{1}{30}^3 cdot frac{29}{30}^2+ frac{1}{30}^4 cdot frac{29}{30}+ frac{1}{30}^5$, approximately $0.104%$

The answer for the second question is simply J.'s probability multiplied by the number of students, approcimately $3.12%$

We had a debate with a fellow student about this explanation, but couldn't convince each other. Is my explanation correct, or if not, where did I make a mistake?

Regarding the first question:

This is the kind of situation in which a binomial distribution would occur: if there's $p$ probability of an event occurring one of $r$ times in a sequence, you need to account for the number of sequences that also have that event occurring.

For example, say I flip a coin $5$ times: how many times does it pop up heads $2$ times? Well, first, how many sequences of $5$ flips have $2$ heads appearing? That's $binom 5 2$. From there, the probability of $2$ successes is the probability $p$ of flipping heads, to the power of $2$, times the probability $1-p$ of instead flipping tails, to the power of $5-2=3$.

More concretely, the probability is given by

$$P(text{2 heads in 5 flips}) = binom 5 2 p^2(1-p)^3$$

Note that I did not explicitly say $p=1/2$ for a reason, that being maybe I'm not flipping a fair coin. My motivation for doing this is because I don't want you to get some wires crossed in your head with how the calculation is performed, so I want to reiterate - my coin is not necessarily fair in this scenario.

Okay, so let's say I want at most $2$ heads. We can simply generalize the above: find the probability of $0,1,$ and $2$ heads; i.e.

$$begin{align}
P(text{at most 2 heads}) &= P(text{0 heads}) + P(text{1 heads}) + P(text{2 heads})\
&= binom 5 0 p^0(1-p)^5 + binom 5 1 p^1(1-p)^4 + binom 5 2 p^2(1-p)^3\
&= sum_{k=0}^2 binom 5 k p^k(1-p)^{5-k}
end{align}$$

In general: say we have a sequence of $n$ trials, all independent from each other. We define the outcome of each to be a success or failure, with probabilities $p$ and $1-p$ respectively. Then the binomial distribution says that the probability of $r$ successes (where $r$ is an integer from $0$ to $n$) is given by

$$P(r ; text{successes}) = binom n r p^r (1-p)^{n-r}$$

If we want "at most" some number of successes, at least, or whatever, we simply sum the above repeatedly, using whichever $r$ values are the most applicable.

In your scenario, OP, $n=5, p=1/30.$ In the first question, we want $rgeq 2$ (so sum over $r=2,3,4,5$).

If you want to look into this topic further, Khan Academy is probably a good resource to start. Here's a link.

The second question is more benign than the above binomial distribution problem, in my opinion. Consider the complementary approach - what is the probability of not having a repeat? There are $30(29)(28)(27)(26)$ ways of choosing students to avoid repeats; divide that by the total number of possible student selections, $30^5$, and that gives you the probability of not having a repeat student. Thus the probability of a repeat is $1$ minus that.

Your first part is not quite right, since you haven't accounted for the fact that $J$ can present solutions in different pairs (or triples, or quads) of classes. Thus the probability that $J$ is selected twice should be
$$
binom{5}{2}frac{29*28*27}{30^5}+binom{5}{3}frac{29*28}{30^5}+binom{5}{4}frac{29}{30^5}+binom{5}{5}frac{1}{30^5}approx.00936
$$

There are $30cdot 29cdot28cdot27cdot26$ ways of choosing students to share without repetitions, out of a total of $30^5$ total ways of choosing students with no restrictions. Thus the probability of having a restriction is
$$
1-frac{30cdot 29cdot28cdot27cdot26}{30^5}approx .296
$$

Edit: Note that $.296$ is NOT simply $.00936*30approx.281$ as you had suspected; this is because it is possible (albeit unlikely) that two pairs of students present twice!

Your first calculation is incorrect, because you haven't considered that there are multiple ways that J. can be chosen twice, for example. $left({1over30}right)^2left({29over30}right)^2$ is the probability that J. is called on in both history and English, say, but there are ${5choose2}$ ways to choose the two classes that he is called on in. Similarly, for three or four classes.

As to the second question, you have problems with double counting. When you add add up the probabilities to get J. is called on twice, or Susan is called on twice, or ..., you have to worry about the possibility that each is called on twice. You've counted that twice. The best way to do the second part is as $1$ minus the probability that $5$ different students are called on:
$$1-5!{30choose5}left({1over30}right)^5$$

I'd approach both questions backwards. The probability that J is not called at all is $(29/30)^5$ and the probability that J is called exactly once is $5(1/30)(29/30)^4$. So the probability that J is called $0$ or $1$ time is approximately $0.9896$, making the probability that J will be called at least twice approximately $0.0104$.

As for the second question, there are $30^5$ ways of selecting a student from each of $5$ classes and there are $30 cdot 29 cdot 28 cdot 27 cdot 26$ ways to choose five students without repeating any. So the probability that no one gets called twice is $0.7037$ and the probability that at least one student is called twice is $0.2963$.