## anonymous 5 years ago 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

1. anonymous

ur in college?

2. anonymous

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

3. anonymous

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

4. anonymous

do you have answers to confirm?

5. 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.

6. anonymous

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

7. anonymous

is it with replacement or without? lol

8. anonymous

What do you think?

9. anonymous

i guess without :P

10. anonymous

yeah i always get confused with statistics

11. anonymous

you may have to use hypergeometric forumula

12. anonymous

wwhat i wrote is binomial

13. 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,

14. anonymous

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

15. anonymous

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

16. anonymous

I don't have answers to go with this.

17. anonymous

anyone figure this out?

18. anonymous

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

19. anonymous

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

20. anonymous

Proof?

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. 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.

27. anonymous

for which part?

28. anonymous

even though probability of getting boy or girl is different?

29. anonymous

oh wait you already accounted for that lol

30. 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.

31. anonymous

I just came back from dinner lol

32. anonymous

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

33. anonymous

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

34. anonymous

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

35. anonymous

a?

36. anonymous

Good job.

37. anonymous

ok so what did u find out so far?

38. anonymous

Nothing :(

39. anonymous

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!

40. anonymous

You solved it first :(

41. anonymous

D) ?

42. anonymous

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

43. anonymous

Very funny, Mr; there is no D.

44. anonymous

lol b I meant

45. anonymous

is that correct though?

46. anonymous

That equal 3/10 - does that seem correct?

47. anonymous

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

48. anonymous

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

49. anonymous

what do u think then?

50. anonymous

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

51. anonymous

what is it

52. anonymous

i will look at it soon, im tied up atm

53. 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.

54. anonymous

Their names* UGGGGGGGGGH grammar fail.

55. anonymous

6/720?

56. anonymous

¬_¬ maybe