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Mathematics 4 Online
OpenStudy (anonymous):

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?

OpenStudy (anonymous):

I need just the setup of how to get the answer

OpenStudy (anonymous):

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

OpenStudy (anonymous):

(5c10+2c10)/7c20

OpenStudy (anonymous):

Is that right

OpenStudy (anonymous):

the c part is throwing me off

OpenStudy (anonymous):

are they asking you for a formula? I don't do formulas, but I can tell you how to go about finding the answer

OpenStudy (anonymous):

I just need to be shown how to get the answer. I'm not sure. I'm just guessing on what I've seen.

OpenStudy (anonymous):

Ok let's walk through it

OpenStudy (anonymous):

since there are 10 men and 10 women (20 people total), what is the probability of picking a woman out of the bunch?

OpenStudy (anonymous):

going at it one by one helps you with the big picture

OpenStudy (anonymous):

2

OpenStudy (anonymous):

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

OpenStudy (anonymous):

get it?

OpenStudy (anonymous):

Ok I see

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so the probability of picking the second woman is the number of women left divided by the number of people left

OpenStudy (anonymous):

what would that be?

OpenStudy (anonymous):

19/20

OpenStudy (anonymous):

nope. how many women are left?

OpenStudy (anonymous):

(after picking 1 woman out)

OpenStudy (anonymous):

9

OpenStudy (anonymous):

good

OpenStudy (anonymous):

how many total people are left?

OpenStudy (anonymous):

(after picking 1 woman out)

OpenStudy (anonymous):

19

OpenStudy (anonymous):

great

OpenStudy (anonymous):

since there are only 9 women left, out of a total 19 people left, the probability of picking the 2nd woman = 9/19

OpenStudy (anonymous):

get it?

OpenStudy (anonymous):

yes so the probability is 9/19

OpenStudy (anonymous):

based on what we did, what is the probability of picking the 3rd woman?

OpenStudy (anonymous):

(we're going to end up picking 5 women and 2 men, so we're not done)

OpenStudy (anonymous):

what is the probability of picking the 3rd woman?

OpenStudy (anonymous):

9/18

OpenStudy (anonymous):

no

OpenStudy (anonymous):

the total number of people left is correct (18), but the number of women left is not correct

OpenStudy (anonymous):

how many women do we have left after taking out 2 women?

OpenStudy (anonymous):

7

OpenStudy (anonymous):

um, we started with 10 women and we took out 2. how many is left?

OpenStudy (anonymous):

8

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so what is the probability of picking the 4th woman?

OpenStudy (anonymous):

(we have taken out 3 women)

OpenStudy (anonymous):

7

OpenStudy (anonymous):

7 out of?

OpenStudy (anonymous):

17

OpenStudy (anonymous):

good

OpenStudy (anonymous):

and the probability of picking the 5th woman?

OpenStudy (anonymous):

6/16

OpenStudy (anonymous):

great. now we're done with the women. on to the men.

OpenStudy (anonymous):

we have 10 men left, and since we've taken out 5 women, we have a total of 15 people left

OpenStudy (anonymous):

so the probability of picking the first man = 10/15

OpenStudy (anonymous):

what is the probability of picking the 2nd man?

OpenStudy (anonymous):

9/14

OpenStudy (anonymous):

that's correct. so now we know the probability of picking each of the 5 women and 2 men, now let's combine them.

OpenStudy (anonymous):

if you have to pick 5 women AND 2 men, then we get to multiply the probability of picking each of them

OpenStudy (anonymous):

so the probability: 10/20 x 9/19 x 8/18 x 7/17 x 6/16 x 10/15 x 9/14

OpenStudy (anonymous):

so multiply all that together

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

I'm not getting how to multiply all that together with fractions. Do u divided the fractions as low as they can go

OpenStudy (anonymous):

i hate to butt in, but there is a different way of doing this, that might make some sense.

OpenStudy (anonymous):

Can u show me please

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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 \[\dbinom{10}{2}=\frac{10\times 9}{2}\]

OpenStudy (anonymous):

you multiply these numbers together by the counting principle, and put them in the numerator so the very short answer to your equation is \[\frac{\dbinom{10}{5}\times \dbinom{10}{2}}{\dbinom{20}{7}}\]

OpenStudy (anonymous):

*question, not equation

OpenStudy (anonymous):

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

OpenStudy (anonymous):

some calculators have buttons for these, so you can do it that way otherwise you have a ton of cancelling and multiplication to do

OpenStudy (anonymous):

if it was me, i would use wolfram

OpenStudy (anonymous):

what is that

OpenStudy (anonymous):

10/20/7

OpenStudy (anonymous):

how do u divide that when 20/7 want divide down

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@mariapernez11 do you know how to compute \(\dbinom{n}{k}\) in general?

OpenStudy (anonymous):

no I don't

OpenStudy (anonymous):

then you need to learn that to do these problems, that should come first

OpenStudy (anonymous):

\[\left(\begin{matrix}n \\ k\end{matrix}\right)=_nC_k=\frac{n!}{(n-k)!k!}\]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.

OpenStudy (anonymous):

I have two more of these two do. What is the answer to this one, cause I'm so confused on this problem

OpenStudy (anonymous):

I'll use that formula on the next two

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@satellite73 thanks for butting in and introducing wolfram alpha! Hadn't heard of it before and it's wicked cool!

OpenStudy (anonymous):

@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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