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Suppose there are 15 rooms among which a person in present in one room. If i pick 5 rooms out of them, what is the probability that the person is present in one of the 5 rooms selected?

No of ways we can pick 5 rooms: 15C5, So answer could be x/15C5

Where am i stuck is how to figure out x? Among 15C5 selections, which of them will contain that specific room in which the person is there?

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  • $\begingroup$ Please show what you have tried and where you are stuck. $\endgroup$ – Gummy bears Sep 23 '15 at 4:48
  • $\begingroup$ Wouldn't this just be $\frac{1}{3}$, since we are splitting up the 15 total rooms into three sets of 5 rooms, and then seeing if the person is in one of the three sets? I feel like I'm missing something though... $\endgroup$ – Brevan Ellefsen Sep 23 '15 at 4:51
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    $\begingroup$ @BrevanEllefsen Nope. The five rooms aren't fixed. It's not like there are three groups of five to chose from. We have tho chose all the five members of the group. $\endgroup$ – Gummy bears Sep 23 '15 at 5:13
  • $\begingroup$ @Gummybears oh ya, combinations... duh. I really dropped the ball on that one. Time to take another probability course... been way too focused on real analysis.... $\endgroup$ – Brevan Ellefsen Sep 23 '15 at 5:19
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Hint

There is only one way to choose the occupied room, what about the rest ?

Answer:

Why confused ? $ Pr = {1\choose 1}{14\choose4} / {15\choose 5}$

Of course, you can leave out the ${1\choose 1}$ , but it shows exactly what you did .

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First, chose the single door that has the person behind it. Then, the four remaining doors can be chosen in any fashion.

Basically, how will you chose 4 doors from a total of 14 doors (as we have already chosen the one with the person behind it)?

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