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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 $\frac{72}{180}\cdot \frac{71}{179}\cdot\dots$ ?

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  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)

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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

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

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  • $\begingroup$ 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. $\endgroup$ – MikhaelM Jan 30 '17 at 7:23
  • $\begingroup$ See my edited answer. $\endgroup$ – Kanwaljit Singh Jan 30 '17 at 7:31
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    $\begingroup$ And if you choose combination you not need to take care about arrangements (which one is picked first , second and so on). $\endgroup$ – Kanwaljit Singh Jan 30 '17 at 7:33
  • $\begingroup$ 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? $\endgroup$ – MikhaelM Jan 30 '17 at 7:42
  • $\begingroup$ For 1 windows we have 1 from window × 14 not. We have 2 from windows × 13 not. $\endgroup$ – Kanwaljit Singh Jan 30 '17 at 8:13

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