# Probability of students having OS when picked at random

Out of $180$ students, $72$ have Windows, $54$ have Linux, $36$ have both Windows and Linux and the rest ($18$) have OS X. What's the probability that out of $15$ randomly picked students:

i) At most two won't have Windows;

ii) At least one will have OS X.

So for the first point I just thought that the probability is the sum of probabilities that "none will have Windows" plus "one will have Windows" plus "two will have windows". Is this a correct train of thought? How do I calculate the probabilities though?

If it were just $1$ student picked, the probability that the student didn't have windows would be $\left(\frac{72}{180}\right)$, right? But I pick $15$, is it $\frac{72}{180}\cdot \frac{71}{179}\cdot\dots$ ?

## 2 Answers

1. For your first question, it will be the sum of probabilities that no one has windows, exactly one student doesn't have windows, and exactly two students don't have windows.

2. P=1- P(no one has X)

Hint -

Case 1 -

Sum of probabilities that no one has windows, exactly one and exactly two students don't have windows.

Case 2 -

At least one have OS X = 1 - No one have OS X

$= 1 - \frac{\binom{162}{15}}{\binom{180}{15}}$

• What about the probability that No one will have OS X? Will that just be the product $\frac{162}{180}\cdot \frac{161}{179}\cdot ...\cdot \frac{148}{166}$ ? Because this seems a little odd to me. – MikhaelM Jan 30 '17 at 7:23
• See my edited answer. – Kanwaljit Singh Jan 30 '17 at 7:31
• And if you choose combination you not need to take care about arrangements (which one is picked first , second and so on). – Kanwaljit Singh Jan 30 '17 at 7:33
• Oh, ok, so for my first case I'd have the sum of probabilities that none have windows, exactly 1 has windows and exactly two have windows. I know that none to have windows is $\frac{\binom{72}{15}}{\binom{180}{15}}$ but what about exactly one/two to have windows? How do I compute that? – MikhaelM Jan 30 '17 at 7:42
• For 1 windows we have 1 from window × 14 not. We have 2 from windows × 13 not. – Kanwaljit Singh Jan 30 '17 at 8:13