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probability (simple) (1 Viewer)

Eagles

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ok.. I can't think tonight.

10 competitors, 4 chosen randomly for drug tests, 3 of them are Japanese

whats the probability that the sample contains at least one Japanese competitior?

Answer: .8333
(how'd they get that)

sample space = 10C4 = 210....
 
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acmilan

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Another equivalent way:

3C1*7C3 + 3C2*7C2 + 3C3*7C1 = 105 + 63 + 7 = 175

Divide by 210 to get .8333
 

Eagles

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acmilan said:
Another equivalent way:

3C1*7C3 + 3C2*7C2 + 3C3*7C1 = 105 + 63 + 7 = 175
what did you do? (idk how I survived prob in highschool..)
 

acmilan

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Theres 3 japanese, to have at least one you can have one, two or three. If you have 1 japanese, then 3 will be not japanese, ie 3C1*7C3. Similarly 2 japanese and 2 not japanese: 3C2*7C2 and 3 japanese and 1 not is 3C3*7C1. Add these together and divide by the total outcomes for the probability
 

boongsta

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ouch my head hurts trying to understand wat u guys are talking about ... im guessing those formulas aren't in 2 unit maths ay ... dont think i seen them before
 

SaHbEeWaH

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yeah they're using 3u method but it's basically the same thing

using 2u is much simpler to look at

P(at least one japanese) = 1 - P(no japanese)
= 1 - (7/10 * 6/9 * 5/8 * 4/7) = 0.8333...
 

ephemeral

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It uses something called the Hypergeometric Distribution. You should probably look it up because it's really simple and is applicable in many questions.

so written in that form it is 1- ((7C4.3C0)/(10C4)) which comes out as .83333 as above.

Basically, for any question like this, you have 2 groups of things (japanese and non-japanese) and just put in the numbers of each you want over the total number of combinations.
 

Kutay

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hey i do the 3unit way as well but i was wondering can you tackle this question by the 2unit way as for this question dont understnd how the 3unit way works
 

word.

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Kutay said:
hey i do the 3unit way as well but i was wondering can you tackle this question by the 2unit way as for this question dont understnd how the 3unit way works
here is your answer:

SaHbEeWaH said:
yeah they're using 3u method but it's basically the same thing

using 2u is much simpler to look at

P(at least one japanese) = 1 - P(no japanese)
= 1 - (7/10 * 6/9 * 5/8 * 4/7) = 0.8333...
 

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