The scales forces of a business consists of 10 men and 10 women. A production unit of 7 people is set up at random. What is the probability that will consists of 5 men and 2 women? The probability is?
I need just the setup of how to get the answer
first, state the probability of getting 5 men out of 20 total people then state the probability of getting 2 women out of 20 total people since we're trying to find the probability of 5 men AND 2 women, "AND" in probability language is a multiplication
(5c10+2c10)/7c20
Is that right
the c part is throwing me off
are they asking you for a formula? I don't do formulas, but I can tell you how to go about finding the answer
I just need to be shown how to get the answer. I'm not sure. I'm just guessing on what I've seen.
Ok let's walk through it
since there are 10 men and 10 women (20 people total), what is the probability of picking a woman out of the bunch?
going at it one by one helps you with the big picture
2
no. there are 10 women, 10 men. the probability of picking a woman is the number of women divided by the total number of people = 10/20 = 1/2
get it?
Ok I see
alright so after i pick the first woman, out of 20 people, you have 19 left. 9 of them are women, 10 of them are men
so the probability of picking the second woman is the number of women left divided by the number of people left
what would that be?
19/20
nope. how many women are left?
(after picking 1 woman out)
9
good
how many total people are left?
(after picking 1 woman out)
19
great
since there are only 9 women left, out of a total 19 people left, the probability of picking the 2nd woman = 9/19
get it?
yes so the probability is 9/19
based on what we did, what is the probability of picking the 3rd woman?
(we're going to end up picking 5 women and 2 men, so we're not done)
what is the probability of picking the 3rd woman?
9/18
no
the total number of people left is correct (18), but the number of women left is not correct
how many women do we have left after taking out 2 women?
7
um, we started with 10 women and we took out 2. how many is left?
8
yes, the probability of picking the 3rd woman is 8 (the number of women left) divided by 18 (the total number of people left) = 8/18
so what is the probability of picking the 4th woman?
(we have taken out 3 women)
7
7 out of?
17
good
and the probability of picking the 5th woman?
6/16
great. now we're done with the women. on to the men.
we have 10 men left, and since we've taken out 5 women, we have a total of 15 people left
so the probability of picking the first man = 10/15
what is the probability of picking the 2nd man?
9/14
that's correct. so now we know the probability of picking each of the 5 women and 2 men, now let's combine them.
if you have to pick 5 women AND 2 men, then we get to multiply the probability of picking each of them
so the probability: 10/20 x 9/19 x 8/18 x 7/17 x 6/16 x 10/15 x 9/14
so multiply all that together
yes
I'm not getting how to multiply all that together with fractions. Do u divided the fractions as low as they can go
i hate to butt in, but there is a different way of doing this, that might make some sense.
Can u show me please
i will write it and the reasoning behind it, and you can see which method you prefer you are going to pick 7 out of 20 the number of ways to pick 7 out of twenty is called "20 choose 7" written in notation as \[\dbinom{20}{7}\] and computed via \[\dbinom{20}{7}=\frac{20\times 19\times 18\times 17\times 16\times 15\times 14}{7\times 6\times 5\times 4\times 3\times 2}\] that will be your denominator cancel first, multiply last or use a calculator
now of those 7 you want 5 men and 2 women the number of ways to choose 5 men out of 10 is also called "10 choose 5" written as \[\dbinom{10}{5}\] and computed via \[\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\] similarly the number of ways to choose 2 out of 10 women is \[\dbinom{10}{2}=\frac{10\times 9}{2}\]
you multiply these numbers together by the counting principle, and put them in the numerator so the very short answer to your equation is \[\frac{\dbinom{10}{5}\times \dbinom{10}{2}}{\dbinom{20}{7}}\]
*question, not equation
writing \[\frac{\dbinom{10}{5}\times \dbinom{10}{2}}{\dbinom{20}{7}}\] really just restates the question the numerator is the number of ways to choose 5 out of 10, times the number of ways to choose 2 out of ten the denominator is the number of ways to choose 7 out of 20
some calculators have buttons for these, so you can do it that way otherwise you have a ton of cancelling and multiplication to do
if it was me, i would use wolfram
what is that
10/20/7
how do u divide that when 20/7 want divide down
unfortunately, the very well explained answer given above by @the_lefay is missing a piece you would also need to multiply by another factor to take in to account permutations
@mariapernez11 do you know how to compute \(\dbinom{n}{k}\) in general?
no I don't
then you need to learn that to do these problems, that should come first
\[\left(\begin{matrix}n \\ k\end{matrix}\right)=_nC_k=\frac{n!}{(n-k)!k!}\]And k! = k(k-1)(k-2)...(3)(2)(1) n! = n(n-1)(n-2)...(3)(2)(1) (n-k)! = (n-k)(n-k-1)(n-k-2)...(3)(2)(1) Perhaps.
I have two more of these two do. What is the answer to this one, cause I'm so confused on this problem
I'll use that formula on the next two
here lets ask wolfram to compute the answer to the first one http://www.wolframalpha.com/input/?i=%28%2810+choose+2%29*%2810+choose+5%29%29%2F%2820+choose+7%29
@satellite73 thanks for butting in and introducing wolfram alpha! Hadn't heard of it before and it's wicked cool!
@the_lefay yea for a problem like this, just type it in and out it pops who wants to compute all that nonsense
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