Another probability question (1 Viewer)

[Elise8842]

Hi

Answer at the back doesn't agree with my solution, could anyone explain why I'm wrong and what's the correct way to solve it?

Thanks in advance



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[Kurosaki]

Hi

Answer at the back doesn't agree with my solution, could anyone explain why I'm wrong and what's the correct way to solve it?

Thanks in advance



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Just subtract probability that all of the entrances are chosen.
does that help or would you prefer a full solution?

 

[glittergal96]

Hi

Answer at the back doesn't agree with my solution, could anyone explain why I'm wrong and what's the correct way to solve it?

Thanks in advance



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If they don't use all three entrances then they either:

a) all go in one entrance

or

b) they all go in two entrances.

a) can happen in exactly three ways, one for each entrance.

b) can happen in 3*(2^5-2) ways. First we choose the pair of entrances, 3C2=3. Then we choose a subset of a five element set that is not empty or the full set to go into the first entrance. So we just subtract those two possibilities from the total number of subsets of a 5 element set.

(3+3*(2^5-2))/(3^5)=31/81.