Ten students are applying for 3 positions on a team. the students include 4 boys (adam, alex, anthony, and arnold) and 6 girls (abbey, aurora, agnes, alice, amanda, and anna). all the students have an equal chance of being selected. find the probability that the students selected will include: a) at most 1 girl b) adam, anthony, and alice c) agnes and 2 other students

- anonymous

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- anonymous

ur in college?

- anonymous

naww not in college lol :P highschool why'd u think im in college?

- M

the first one should be (4/10)^3 + (6/10)^1 x (4/10)^2

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- M

do you have answers to confirm?

- anonymous

i was working on his question until you answered it. i was simply wondering why theis person thought i was in college. he asked me a question, i was answering.

- M

second one should be (6/10)^1 x (4/10)^2

- M

is it with replacement or without? lol

- anonymous

What do you think?

- M

i guess without :P

- M

yeah i always get confused with statistics

- M

you may have to use hypergeometric forumula

- M

wwhat i wrote is binomial

- anonymous

Well, you certainly don't 'have to' - if one really wanted to they could solve this by just drawing a (massive) tree diagram, but I think we can be slightly more clever about it,

- M

hmm maybe (6/10)(4/9)(3/8) x 3 + (4/10)(3/9)(2/8)?

- M

crap i'm so confused with this i should just stop lol. gl with the problem.

- anonymous

I don't have answers to go with this.

- anonymous

anyone figure this out?

- anonymous

Don't worry, you can just draw a tree-diagram with 720 branches and be sure of the answer.

- M

[(6/10)(4/9)(3/8) + (4/10)(6/9)(3/8) + (4/10)(3/9)(6/8)] + (4/10)(3/9)(2/8)

- anonymous

Proof?

- M

wait the ordering doesn't matter so you have to divide the first part by 3? or 6?

- anonymous

I preferred it in the old form you posted:
(6/10)(4/9)(3/8) x 3 + (4/10)(3/9)(2/8)

- M

only problem with that is you don't know girl is selected before boys

- anonymous

You don't. But your answer there accounts for it by multiplying by three.

- M

i don't know stat well enough to figure this out without final answer unfortunatley

- anonymous

OK.
If there are three boys chosen, the chance is (clearly... I hope) (4/10)(3/9)(2/8)
The only other option is 1 girl and 2 boys.
This can either be GBB, BGB, BBG, but the chance of each is still (6x4x3)/(10x8x9).
So your answer
\[\frac{(6 \times 4 \times 3) \times 3 + 4 \times 3 \times 2}{720} \]
is right.

- anonymous

for which part?

- M

even though probability of getting boy or girl is different?

- M

oh wait you already accounted for that lol

- anonymous

Yes; there's still a total number of 10*9*8 = 720 in the denominator; the order of the multiplication above is irrelevant.
@ azntiger627 - FACEPALM - if you honestly have no idea which part this is then you have clearly done nothing on this work, and do not deserve help until at least attempt it.

- anonymous

I just came back from dinner lol

- anonymous

Wasting time eating... when you have Maths to do!? That's a cardinal sin in my book.

- anonymous

lol sorry, now I'm rdy, so which part were u guys talking about?

- anonymous

I'll tell you. But you have to guess first.

- anonymous

a?

- anonymous

Good job.

- anonymous

ok so what did u find out so far?

- anonymous

Nothing :(

- M

Newton already solved part (a) for you above. but i think you should go over permutations and combinations before trying to solve the question above. good luck!

- anonymous

You solved it first :(

- anonymous

D) ?

- anonymous

for d, i'm guessing (4C2)(6C1) / 10C3

- anonymous

Very funny, Mr; there is no D.

- anonymous

lol b I meant

- anonymous

is that correct though?

- anonymous

That equal 3/10 - does that seem correct?

- anonymous

yea I got 3/10, is it correct though?

- anonymous

There are 720 possibilities, and you think 216 contain the same three people? (they don't)

- anonymous

what do u think then?

- anonymous

I don't 'think' anything; I know what the answer is.

- anonymous

what is it

- anonymous

i will look at it soon, im tied up atm

- anonymous

There names all begin with A, which is annoying, so just called them ABC.
There are 6 ways of arranging them:
ABC
ACB
etc.
And 10 x 9 x 8 possible ways in total. Continue from there.

- anonymous

Their names* UGGGGGGGGGH grammar fail.

- anonymous

6/720?

- anonymous

¬_¬ maybe

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