probability (simple) (1 Viewer)

[Eagles]

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....

 

[Eagles]

wait... got it.. thanks

(10C4 - 7C4) / 10C4

 

[rama_v]

1-p(no japanese) = 1 - (7C4 / 10C4) = 0.8333 = 5/6

 

[acmilan]

Another equivalent way:

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

Divide by 210 to get .8333

 

[Eagles]

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]

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

 

[Eagles]

Ooooooh I understand now...

cheers all!

 

[boongsta]

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]

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]

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]

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.]

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:

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...