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

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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

Mathematics
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ur in college?
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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Other answers:

  • M
do you have answers to confirm?
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
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
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.
I don't have answers to go with this.
anyone figure this out?
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)
Proof?
  • M
wait the ordering doesn't matter so you have to divide the first part by 3? or 6?
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
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
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.
for which part?
  • M
even though probability of getting boy or girl is different?
  • M
oh wait you already accounted for that lol
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.
I just came back from dinner lol
Wasting time eating... when you have Maths to do!? That's a cardinal sin in my book.
lol sorry, now I'm rdy, so which part were u guys talking about?
I'll tell you. But you have to guess first.
a?
Good job.
ok so what did u find out so far?
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!
You solved it first :(
D) ?
for d, i'm guessing (4C2)(6C1) / 10C3
Very funny, Mr; there is no D.
lol b I meant
is that correct though?
That equal 3/10 - does that seem correct?
yea I got 3/10, is it correct though?
There are 720 possibilities, and you think 216 contain the same three people? (they don't)
what do u think then?
I don't 'think' anything; I know what the answer is.
what is it
i will look at it soon, im tied up atm
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.
Their names* UGGGGGGGGGH grammar fail.
6/720?
¬_¬ maybe

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