Here's the question you clicked on:
xEnOnn
Suppose past data has shown that 60% of the students attend lectures. There are 50 students enrolled in a course. To check attendance in each lecture, 15 names of the students are randomly selected. What is the probability that there are strictly fewer than 5 students being present in a lecture?
Okay, so we want the probability that of the 15 chosen, 4 or less are present.
60% of 50 students is 30 students who attend. And 20 don't attend. Pick the first person, there is a 30/50 chance he is present. Pick the second person, there is a 29/49 chance is is present. Third, 28/48. Fourth, 27/47. Now for the fifth person, we want him to be absent, that is a 20/46 chance. Sixth person, 19/45. And you keep playing this game down to 15 choices. Now it turns out that even though I chose to pick the present people first, it has no bearing on the solution. What I mean is, if you chose the absent people first, you will still get the same answer. That's because you will still get exactly the same denominator. And you will get the same numerator but in a different order. Now multiply this answer by 15 choose 4 to get all orders, and you have found the probability that exactly 4 people show up. Do the same thing for 3, and for 2, and 1, and 0, and add all of these together.
Give me a sec, I am reading and trying to understand it.
Would this be hypergeometric distribution?
I calculated and it indeed is! Thanks for your clear explanation! Thanks!! :D