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the question goes:

pretend there are 2 rooms available, and at any given time, there could be 0, 1, or 2 meetings going on overall over the 2 rooms, and each scenario happens at equal chance. what is the probability that you walk into a room, at some point, and that room has a meeting going on?

clarifications:

1) each room can only host up to one meeting
2) one meeting can only happen in one room

I am a noob at stats... i really do not know how to start on this one. if you could explain to me how this one works in the simplest words that'd be really really helpful!

note

this is neither my hw or anything that i can benefit from other than fulfilling my curiosity to learn. i have taken some intro probability classes online but none of them go into problems like this... and having no TA nor professor for help is difficult for me to figure things out myself :(

1 Answers 1

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The answer is $\frac 12$. If there are $2$ meetings, chance $\frac 13$, you will find a meeting. If there is $1$, chance $\frac 13$ you will find a meeting with probability $\frac 12$ for a total chance of $\frac 13 \cdot \frac12=\frac 16$. As these are disjoint, we can add them to get $\frac 12$. You need to study the fact that you multiply independent probabilities to get the chance of the conjunction and add disjoint probabilities to get the total chance of one.

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    why did you multiply 1/3*12 for 2nd scenario? where's 12 coming from? – 2017-01-06
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    I missed \frac, it was supposed to be $\frac 12$ as the chance you pick the room with the meeting. Fixed – 2017-01-06
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    Ah. It makes sense to me now!! Thank you!!! – 2017-01-06
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    There are two meetings but person is adding in one room. And also one meeting in one room. So 1/3 doesn't make sense. – 2017-01-06
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    @KanwaljitSingh: $\frac 13$ is the chance there is one meeting in progress. One room has a meeting, one does not. You randomly choose one of the two rooms. – 2017-01-06