/sci/ - Science & Math » Thread #13678434

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Anonymous Fri 24 Sep 18:26:04 2021 No.13678434 View Reply Original Report

Quoted By: >>13678444 >>13679558

There is a room filled with 25 people. Of those 25 people, 6 are undercover agents.

If you randomly select 4 people from that room, what is the chance that 3 of those selected are undercover agents?

Anonymous

Anonymous Fri 24 Sep 2021 18:27:46 No.13678444 Report

Quoted By: >>13679558

>>13678434
>who was that Bernoulli guy, anyway

Anonymous

Anonymous Fri 24 Sep 2021 23:01:58 No.13679558 Report

Quoted By:

>>13678434
>If you *randomly* select 4 people
So every choice is unique and equiprobable.
There are a total of ${25\choose 4}=\frac{25\cdot24\cdot23\cdot22}{4!} = 25\cdot23\cdot22 = 12650$ ways of choosing people.
The ways of choosing people such that exactly 3 are agents are ${19\choose 1}\cdot{6\choose 3}=19\cdot20=380$ .
So the probability of selecting 3 undercover agents is $\frac{{19\choose 1}\cdot{6\choose 3}}{{25\choose 4}} \approx 0.03$ .
Every problem like this one is the same thing, it's literally just counting.

>>13678444
Could you elaborate? I suppose you don't mean Bernoulli trials because the events are not independent. Can you modify anything so it works or something?

Anonymous

Anonymous Sat 25 Sep 2021 00:00:26 No.13679719 Report

Quoted By:

amogus

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