Question

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

Answer 1

ur in college?

Answer 2

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

Answer 3

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

Answer 4

do you have answers to confirm?

Answer 5

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.

Answer 6

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

Answer 7

is it with replacement or without? lol

Answer 8

What do you think?

Answer 9

i guess without :P

Answer 10

yeah i always get confused with statistics

Answer 11

you may have to use hypergeometric forumula

Answer 12

wwhat i wrote is binomial

Answer 13

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,

Answer 14

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

Answer 15

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

Answer 16

I don't have answers to go with this.

Answer 17

anyone figure this out?

Answer 18

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

Answer 19

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

Answer 20

Proof?

Answer 21

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

Answer 22

I preferred it in the old form you posted:

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

for which part?

Answer 28

even though probability of getting boy or girl is different?

Answer 29

oh wait you already accounted for that lol

Answer 30

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.

Answer 31

I just came back from dinner lol

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

a?

Answer 36

Good job.

Answer 37

ok so what did u find out so far?

Answer 38

Nothing :(

Answer 39

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!

Answer 40

You solved it first :(

Answer 41

D) ?

Answer 42

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

Answer 43

Very funny, Mr; there is no D.

Answer 44

lol b I meant

Answer 45

is that correct though?

Answer 46

That equal 3/10 - does that seem correct?

Answer 47

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

Answer 48

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

Answer 49

what do u think then?

Answer 50

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

Answer 51

what is it

Answer 52

i will look at it soon, im tied up atm

Answer 53

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.

Answer 54

Their names* UGGGGGGGGGH grammar fail.

Answer 55

6/720?

Answer 56

¬_¬ maybe