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Mathematics 14 Online
OpenStudy (dunnitagainn):

If a jury list contains 20 men and 20 women, what is the probability of randomly selecting 12 members that are all men?

OpenStudy (caution0):

50/50 is this a bubble in question??

OpenStudy (dunnitagainn):

if you mean multiple choice no, it is free response?

OpenStudy (dunnitagainn):

would it be 50/50 for all 12 people?

OpenStudy (caution0):

cant really tell you the answer straight out.. im trying to figure out a way to explain it

OpenStudy (caution0):

2/3 of a jury of 12 = 8 men. There will always be at least 8 men on the jury unless all 5 women are selected, in that case there will be 7 men and 5 women.

OpenStudy (zarkon):

have you done permutations/combinations

OpenStudy (dunnitagainn):

why does there have to be at least 2/3 of the jury as men?

OpenStudy (caution0):

455 ways to select a jury of 12 from a pool of 15 1 way to select all 5 women 120 ways to select 7 men from a pool of 10 120 ways to select a jury with fewer than 8 men

OpenStudy (caution0):

(455-120)/455 = 335/455 = 67/91

OpenStudy (caution0):

does that help??

OpenStudy (dunnitagainn):

yeah it does thank you

OpenStudy (caution0):

no problem ^.^

OpenStudy (zarkon):

really? that is not even close to the answer

OpenStudy (caution0):

how ???

OpenStudy (zarkon):

why are you talking about 5 and 7 and 8 this is a simple hypergeometric distribution problem

OpenStudy (dunnitagainn):

that was confusing me, i thought i must be doing it wrong

OpenStudy (zarkon):

out of the 20 men choose 12 out of the 40 people choose 12 then divide

OpenStudy (dunnitagainn):

so (12/20) * (12/40) ?

OpenStudy (zarkon):

\[\frac{_{20}C_{12}}{_{40}C_{12}}\]

OpenStudy (caution0):

well ig i was doin it wrong.. looks like he got it

OpenStudy (zarkon):

the probability will be a very small number

OpenStudy (dunnitagainn):

Im not sure what the c is? we have not learned that

OpenStudy (zarkon):

ok look at the number of choices you have ... when choosing the men you have 20 choices for the 1st guy, 19 for the sec....as so on.. this gives \[20\times 19\times\cdots\times 9\] number of ways the pick the guys

OpenStudy (zarkon):

then the number of ways to pick 12 people from 12 is \[40\times 39\times\cdots\times 29\]

OpenStudy (zarkon):

then divide \[\frac{20\times 19\times\cdots\times 9}{40\times 39\times\cdots\times 29}\]

OpenStudy (dunnitagainn):

okay, thats what i did earlier, but i thought i was doing it wrong because it was so many steps, thank you so much!!!

OpenStudy (zarkon):

np

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