1

Suppose that in a class of 50 students, the instructor calls on students to answer questions, for each question selecting one student at random and without consideration of whom he has called on in the past. If the instructor asks 20 questions of the class, what is the expected number of students he would have called on?

wouldn't this just be 1 + (49/50) + (48/50) until the twentieth term?

fatwalrus
  • 701
  • Not exactly... because if he calls on the same student the first two times, for example, the probability of calling on a new student will then be $49/50$, not $48/50$. – mjqxxxx Mar 13 '15 at 00:11
  • 8
    The probability that a particular student is ever called on is $1 - (49/50)^{20}$. So the expected number of students called on is just $50\times(1 - (49/50)^{20})\approx 16.62$. – mjqxxxx Mar 13 '15 at 00:14

2 Answers2

1

Have you learned Markov Chains? It is very simple to do it using that. Assume that we have asked one question (thus identifying one unique student).

We can define the Transition Matrix as:

\begin{align} X= \begin{pmatrix} 1/50 & 49/50\\ & 2/50 & 48/50\\ & & \ddots\\ & & & \ddots\\ & & & & 20/50 \end{pmatrix} \end{align}

Let $P=X^{19}$. These are the number of transitions till the end of the lecture (Realize that our initial state was when we had already asked one question).

The answer to your question can then be given by $$\sum_{i=1}^{20} i\times P_{1i}=16.6196$$ Where $P_{1i}$ implies the element in the first row and $i^{th}$ column.

Nitish
  • 1,184
  • I know that the original question was to find the mean, but it seems worth emphasizing that this elegant answer also provides the distribution of the number of students called upon. – BruceET Mar 13 '15 at 16:25
  • For those interested in details of Markov chains, the full transition matrix is 50x50. There are 49 transient classes (since one can never transition to a lower state) and state 50 is absorbing (if questioning continues indefinitely each of the 50 students will eventually be called upon). This coordinates with other recent questions on this site about reducible transition matrices. (A further question would be the average number of questions required until all 50 students have participated.) – BruceET Mar 13 '15 at 20:08
  • @BruceTrumbo "A further question would be the average number of questions required until all 50 students have participated." To which the answer is well known. – Did Mar 14 '15 at 12:46
  • @Nitish How did you compute the sum? By computing the matrix P=X^19? – Did Mar 14 '15 at 12:48
  • @Did, Yes. I'm sure there are smarter ways but this was quick and clean. – Nitish Mar 14 '15 at 14:25
  • How did you compute X^19? – Did Mar 14 '15 at 14:32
  • I put the entire 50x50 matrix into R, found its 19th power, and used the first row to verify 16.6196 is the mean and to find 1.427 is the SD. (I suppose there is a way that requires less brute force.) – BruceET Mar 14 '15 at 19:38
  • "I suppose there is a way that requires less brute force." Quite so! And it is mentioned on this page. (Unrelated: Please use @.) – Did Mar 15 '15 at 14:52
1

I'm converting mjqxxxx's comment into a community wiki answer since the only existing answer is astronomical overkill for the question at hand.

The probability that a particular student is ever called on is $1−\left(\frac{49}{50}\right)^{20}$, so [by linearity of expectation] the expected number of students called on is just $50\cdot(1-\left(\frac{49}{50}\right)^{20}\approx16.62$.

joriki
  • 242,601