willa:

Data Management: A jury of 12 people is to be chosen randomly from a list containing 10 men and 12 women. What is the probability that there is a least one man on the jury?

4 weeks ago
Vocaloid:

would be easier to find the probability of getting no men on the jury, then taking 1 - that probability so getting all women, we have 12 women out of a possible 22 people so P(only women) = (12/22)(11/21)(10/20)(9/19)(8/18)(7/17)(6/16)(5/15)(4/14)(3/13)(2/12)(1/11) then P(at least one man) = 1 - P(only women)

4 weeks ago
willa:

so i would divide all those in the brackets then times them to find my answer?

4 weeks ago
kittybasil:

Limited statistics memory, but I believe you multiply first before dividing...?

4 weeks ago
Vocaloid:

yeah it's just a series of fractions so you'd multiply across all the numerators to get 12 factorial, then 22 * 21 * 20... *11 for the denominator, then divide first quantity/second quantity

4 weeks ago
Vocaloid:

i'm sure there's an easier way to express this w/ factorial notation, it's just slipping my mind at the moment

4 weeks ago
willa:

oh ok thanks guys ill try it out and see if get it!!

4 weeks ago
Vocaloid:

for large calculations I would just chuck it into wolframalpha.com and let it simplify the fraction

4 weeks ago
kittybasil:

or symbolab.com

4 weeks ago
willa:

oh ok, never heard of these sites thanks ill use them rn to make it easier, again thanks guys!!<3

4 weeks ago
kittybasil:

4 weeks ago
willa:

hey uh for the part when you said "yeah it's just a series of fractions so you'd multiply across all the numerators to get 12 factorial, then 22 * 21 * 20... *11 for the denominator, then divide first quantity/second quantity" i understand the numerator part but can you explain the denominator part a bit more?

4 weeks ago
Vocaloid:

like, if you have (A/B) * (C/D) you'd multiply the numerators (A*C) and then the denominators (B*D), then the final result is (A*C)/(B*D) same logic with (12/22)(11/21)(10/20)(9/19)(8/18)(7/17)(6/16)(5/15)(4/14)(3/13)(2/12)(1/11) for the numerators multiply 12 * 11 * 10... etc. for the denominators multiply 22 * 21 * 20 * 19 * 18... * 11

4 weeks ago
willa:

i got 3.0974447e+14 idek what the e is lol please help

4 weeks ago
kittybasil:

that seems like \[3.0974447\times10^{-14}\]

4 weeks ago
Vocaloid:

e is an exponent setting my calculator in fraction mode gives me 646645/646646 as the sol'n

4 weeks ago
kittybasil:

You can use "e(numerical value)" instead of "\(10^{\text{numerical value}}\)"

4 weeks ago
kittybasil:

Sorry, that should be \(10^{14}\) not negative 14. My bad

4 weeks ago
willa:

so when it comes down to what the fraction should be would it be " 12!/3.0974447e14 " ??

4 weeks ago
Vocaloid:

product is written as 1 - (12/22)(11/21)(10/20)(9/19)(8/18)(7/17)(6/16)(5/15)(4/14)(3/13)(2/12)(1/11) which can be written as 1 - (12!)/(22*21*20*19*18*17*16*15*14*13*12*11) which simplifies into 646645/646646

4 weeks ago
willa:

so 1-1? 0 is the answer?

4 weeks ago
willa:

@Vocaloid

4 weeks ago
Vocaloid:

"1 - (12!)/(22*21*20*19*18*17*16*15*14*13*12*11) which simplifies into 646645/646646" indicating that 646645/646646 is the solution.

4 weeks ago
willa:

sorry i keep asking repetitive questions but i know you said that is the solution but i gotta find the probabilty as in a percentage or fraction so when i divided those two numbers i got 99%.. is that what my answer should be or am i way off?

4 weeks ago
Vocaloid:

"percentage or fraction" seems to imply that 646645/646646 would be an acceptable solution however, if you must report the solution as a percentage, then yes, 99% would be an acceptable solution do they specify how many digits to round to?

4 weeks ago
willa:

no they dont the question i posted above which was "A jury of 12 people is to be chosen randomly from a list containing 10 men and 12 women. What is the probability that there is a least one man on the jury?" was all they asked

4 weeks ago
Vocaloid:

hm. 646645/646646 is more accurate, 99% could work too

4 weeks ago
willa:

alright thanks so much for the help i really appreciate it!!

4 weeks ago
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