## perl 2 years ago A box contains 8 dark chocolates, 8 white chocolates, and 8 milk chocolates. I choose chocolates at random (yes, without replacement; I’m eating them). What is the chance that I have chosen 20 chocolates and still haven’t got all the dark ones? what is your answer to that

1. KABRIC

dont really get it

2. UsukiDoll

choose milk chocolates choose white chocolates and just 4 dark chocolates XD

3. Chelsea04

the question says that you still haven't got ALL the dark ones, so you COULD get 4 dark chocolates to 7 dark chocolates, but not the full 8

4. perl

oh woops

5. perl

i thought it said 'any'

6. Chelsea04

do you get the question now?

7. perl

yes

8. UsukiDoll

I'm thinking milk white milk white milk white milk white milk white

9. perl

you wantthe probability of not getting all 8 dark chocolates. you are guaranteed at least 4 dark chocolates so the possibilities are 4 dark, 5 dark, 6 dark, 7 dark

10. Chelsea04

yep, so if you find the probability of getting 8 dark chocolates, you can minus it from 1 and you'll get the probability of getting 4 dark, ... 7 dark

11. perl

there are 4 favorable cases, out of 5 total possible cases favorable cases : : exactly 4 dark , 16 non-dark : exactly 5 dark 15 non-dark exactly 6 dark 14 non-dark exactly 7 dark 13 non-dark total cases : exactly 4 dark , 16 non-dark : exactly 5 dark 15 non-dark exactly 6 dark 14 non-dark exactly 7 dark 13 non-dark exactly 8 dark , 12 non-dark I see no reason not to treat these choices as equally likely. therefore the probability of not picking all dark ones is 4/5

12. perl

since they must add up to 20 , these are the only possibilites

13. perl

notice 3 dark 17 non-dark is impossible

14. perl

the situation is constrained by two facts there must be no more than 16 non-dark chocolates there must be at least 4 dark chocalates

15. perl

that produces (16,4) (15, 5 ) (14,6) (13, 7) (12,8)

16. perl

the situation is constrained by two facts you cannot have more than 16 non-dark chocolates you cannot have more than 8 dark chocolates there must be at least 4 dark chocalates so the possibilities are (# non-dark, # dark) (16,4) (15, 5 ) (14,6) (13, 7) (12,8) there is 4/5 favorable

17. perl

can i get a yes?

18. Chelsea04

sure, i guess

19. perl

lol

20. Chelsea04

the explanation seems reasonable enough

21. perl

i think its more complicated, let X = number of dark ones , Y = number of non-dark chocolate how many ways can you choose 16 non dark and 4 dark? how many ways can you choose 15 non-dark and 5 dark?

22. perl

P( chance not getting all dark in 20 draws) = P( X = 4 & Y= 16 or X= 5 & Y=15 or X=6 & Y=14 or X=7 or Y=13 )

23. Chelsea04

that's what i thought too, but if you add up all the ways to not get all 8 dark chocolate, i'm like 90% sure you'll get 4/5...cause there are like 60 ways if you account for each individual selection

24. perl

P( X = 4 & Y= 16 or X= 5 & Y=15 or X=6 & Y=14 or X=7 or Y=13 ) = P ( X=4 & Y=16) + P(X=5 & Y=15) + ...

25. Chelsea04

btw, what level of maths is this?

26. perl

not sure

27. Chelsea04

i meant like are you still in school or university?

28. perl

P(Y=16 & X=4) = P(Y=16) * P (X=4 | Y = 16) =

29. perl

im in university

30. Chelsea04

oh, then you might not want to listen to me, i'm in yr 11 doing yr 12 maths so yea...

31. perl

i find this problem confusing

32. Chelsea04

whoops, should have told you that at the start!

33. perl

im trying to force these into mutually exclusive possibilities

34. perl

you cannot have more than 16 non-dark chocolates you cannot have more than 8 dark chocolates there must be at least 4 dark chocalates so the possibilities are (# non-dark, # dark) (16,4) (15, 5 ) (14,6) (13, 7) (12,8)

35. perl

but are these equally likely?

36. perl

if these possibilities are equally likely to occur (and they are mutually exclusive), then the probability is 4/5

37. Chelsea04

i'd assume so, unless some were heavier or bigger than the others (which isn't so)

38. perl

im use to working on problems like 5/12*4/11 , those sorts of problems

39. Chelsea04

me too, I guess you could do that, but you'd be there forever!

40. perl

41. Chelsea04

I found the probability of getting all 8 chocolates, kinda took a while and subtracted it from 1

42. perl

can you show me your work

43. perl

44. perl

|dw:1366635152002:dw|

45. Chelsea04

hmm..i guess you could use combinatorics

46. Chelsea04

does order matter?

47. perl

because i want Probability ( 4 dark& 16 non-dark) exactly

48. perl

$\frac{ 8C4*16C16 }{24C20 }+\frac{ 8C5*16C15 }{24C20}+\frac{ 8C6*16C14}{24C20}+\frac{ 8C7*16C13 }{24C20 }$

49. Chelsea04

what do you get from that?

50. perl

the first is the case 4 dark & 16 non-dark then 5 dark& 15 nondark

51. perl

the order of eating them does not count, but we should label in our mind the chocolates. if you want, label the dark chocolates 1-8, and label the non-dark chocolates 9-24

52. perl

so now we have the problem dark chocolate = { 1,2,3,4,5,6,7,8} , non-dark { 9,10,11,12,... 23,24} , the numbers are going to be labels for the chocolates, ok ?

53. perl

i could have used letters,

54. Chelsea04

sure

55. perl

so for instance, suppose we want four dark chocolates, and 16 non-dark chocolate. so we do probability = # favorable outcomes / # total possibilities the favorable, first choose 4 dark chocolates which has a total of 8 choose 4 ways to do it, then multiply by that how many ways can you choose 16 non-dark from 16 non dark. there is only one way. so im using multiplication rule (and probability)

56. perl

the denominator is the total number of ways to choose 20 chocolates , and there are 24 choose 20 ways to choose 20 chocolates

57. perl

anyways, show me your work, and we can compare answers . sorry if my explanation is confusing, im not exactly sure of the correct jargon

58. perl

so I got a probability of 629/759 = .8287

59. Chelsea04

I tried using permutations so like: $\frac{ nPr(16,16)*nPr(8,4)+nPr(16,15)*nPr(8,5)...etc }{ nPr(24,20) }$ but then the amswer was: $\frac{ 5 }{ 245157 }$ which doesn't seem right...AT ALL... so i replaced all the P with C to use combinatorics and i got 629/759, which seems more accurate than permutations

60. Chelsea04

it's close to 4/5 at least

61. perl

thats right, and yes its close to 4/5, but 4/5 would give you a wrong answer. i know how teachers are, lol

62. KABRIC

63. perl

and you could solve it using the complement approach .

64. perl

kabric, how do you know?

65. Chelsea04

yea, basically what i was thinking

66. perl

chelsea, how did you get nCr using pretty print?

67. Chelsea04

i just typed it?

68. perl

$1-\frac{ \left(\begin{matrix}8 \\ 8\end{matrix}\right)*\left(\begin{matrix}16 \\ 12\end{matrix}\right) }{ \left(\begin{matrix}24 \\ 20\end{matrix}\right) }$

69. perl

that is 8C8 , 16 C 12

70. perl

do you how it comes out the same ?

71. Chelsea04

what do you mean?

72. Chelsea04

you sort of lost me there

73. perl

oh ok sorry

74. perl

|dw:1366637463112:dw|

75. Chelsea04

yes...

76. perl

P( X>7 ) = P(X=8 or X=9 or .. ) but X=9 and higher is impossible so we only need to look at X = 8

77. perl

we cant have 9 dark ones, since there arent 9 dark ones

78. Chelsea04

sure

79. perl

ok lets not pursue this further, i have a new question answering, follow me

80. WalterDe