anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
ur in college?
anonymous
  • anonymous
naww not in college lol :P highschool why'd u think im in college?
M
  • M
the first one should be (4/10)^3 + (6/10)^1 x (4/10)^2

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More answers

M
  • M
do you have answers to confirm?
anonymous
  • 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
  • M
second one should be (6/10)^1 x (4/10)^2
M
  • M
is it with replacement or without? lol
anonymous
  • anonymous
What do you think?
M
  • M
i guess without :P
M
  • M
yeah i always get confused with statistics
M
  • M
you may have to use hypergeometric forumula
M
  • M
wwhat i wrote is binomial
anonymous
  • 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
  • M
hmm maybe (6/10)(4/9)(3/8) x 3 + (4/10)(3/9)(2/8)?
M
  • M
crap i'm so confused with this i should just stop lol. gl with the problem.
anonymous
  • anonymous
I don't have answers to go with this.
anonymous
  • anonymous
anyone figure this out?
anonymous
  • anonymous
Don't worry, you can just draw a tree-diagram with 720 branches and be sure of the answer.
M
  • 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
  • anonymous
Proof?
M
  • M
wait the ordering doesn't matter so you have to divide the first part by 3? or 6?
anonymous
  • 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
  • M
only problem with that is you don't know girl is selected before boys
anonymous
  • anonymous
You don't. But your answer there accounts for it by multiplying by three.
M
  • M
i don't know stat well enough to figure this out without final answer unfortunatley
anonymous
  • 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
  • anonymous
for which part?
M
  • M
even though probability of getting boy or girl is different?
M
  • M
oh wait you already accounted for that lol
anonymous
  • 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
  • anonymous
I just came back from dinner lol
anonymous
  • anonymous
Wasting time eating... when you have Maths to do!? That's a cardinal sin in my book.
anonymous
  • anonymous
lol sorry, now I'm rdy, so which part were u guys talking about?
anonymous
  • anonymous
I'll tell you. But you have to guess first.
anonymous
  • anonymous
a?
anonymous
  • anonymous
Good job.
anonymous
  • anonymous
ok so what did u find out so far?
anonymous
  • anonymous
Nothing :(
M
  • 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
  • anonymous
You solved it first :(
anonymous
  • anonymous
D) ?
anonymous
  • anonymous
for d, i'm guessing (4C2)(6C1) / 10C3
anonymous
  • anonymous
Very funny, Mr; there is no D.
anonymous
  • anonymous
lol b I meant
anonymous
  • anonymous
is that correct though?
anonymous
  • anonymous
That equal 3/10 - does that seem correct?
anonymous
  • anonymous
yea I got 3/10, is it correct though?
anonymous
  • anonymous
There are 720 possibilities, and you think 216 contain the same three people? (they don't)
anonymous
  • anonymous
what do u think then?
anonymous
  • anonymous
I don't 'think' anything; I know what the answer is.
anonymous
  • anonymous
what is it
anonymous
  • anonymous
i will look at it soon, im tied up atm
anonymous
  • 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
  • anonymous
Their names* UGGGGGGGGGH grammar fail.
anonymous
  • anonymous
6/720?
anonymous
  • anonymous
¬_¬ maybe

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