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When a teacher asks a question, a student has a probability of 0.4 of being asked. Assume the occurrence is independent. What is the mode of the number of questions raised by the teacher it takes for the same student to be asked 2 questions?

I am stuck at solving this problem. I tried to identify the distribution of the random variable first and I am thinking that it follows a negative binomial distribution, but I am unsure. Please give me hints on how do I proceed with this.

Thank you.

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    $\begingroup$ yes, negative binomial. then, find the probability mass function for that distribution for a few values until you are sure you found the one with highest probability. $\endgroup$ – John L Mar 3 at 16:52
  • $\begingroup$ mode for those not aware what it is like me. :) $\endgroup$ – Easy Points Mar 5 at 0:10
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There are several parameterizations of the negative binomial distribution. So it is always necessary to specify the form of the PDF being used.

If $Y$ is the number of the question on which the student gets asked for the second time, then the PDF of $Y$ is $f(y) = (y-1)(.4)^2(.6)^{y-2},$ for $y = 2, 3, 4, \dots .$ (There will be $y$ questions (Bernoulli trials) altogether. The first time the student is called upon can be at any of the $y-1$ questions before the second time.)

For $y = 2, 3, \dots 10,$ the values of the PDF are shown in the R output below (ignore line numbers in [ ]s.)

y = 2: 10
pdf = (y-1)*.4^2*.6^(y-2)
cbind(y, pdf)
       y        pdf
 [1,]  2 0.16000000
 [2,]  3 0.19200000
 [3,]  4 0.17280000
 [4,]  5 0.13824000
 [5,]  6 0.10368000
 [6,]  7 0.07464960
 [7,]  8 0.05225472
 [8,]  9 0.03583181
 [9,] 10 0.02418647

plot(y, pdf, type="h", lwd=3, ylim=c(0,.2), main="Negative Binomial PDF")
abline(h=0, col="green2")

enter image description here

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