If a jury list contains 20 men and 20 women, what is the probability of randomly selecting 12 members that are all men?
50/50 is this a bubble in question??
if you mean multiple choice no, it is free response?
would it be 50/50 for all 12 people?
cant really tell you the answer straight out.. im trying to figure out a way to explain it
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.
have you done permutations/combinations
why does there have to be at least 2/3 of the jury as men?
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
(455-120)/455 = 335/455 = 67/91
does that help??
yeah it does thank you
no problem ^.^
really? that is not even close to the answer
how ???
why are you talking about 5 and 7 and 8 this is a simple hypergeometric distribution problem
that was confusing me, i thought i must be doing it wrong
out of the 20 men choose 12 out of the 40 people choose 12 then divide
so (12/20) * (12/40) ?
\[\frac{_{20}C_{12}}{_{40}C_{12}}\]
well ig i was doin it wrong.. looks like he got it
the probability will be a very small number
Im not sure what the c is? we have not learned that
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
then the number of ways to pick 12 people from 12 is \[40\times 39\times\cdots\times 29\]
then divide \[\frac{20\times 19\times\cdots\times 9}{40\times 39\times\cdots\times 29}\]
okay, thats what i did earlier, but i thought i was doing it wrong because it was so many steps, thank you so much!!!
np
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