Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. KABRIC
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    dont really get it

  2. UsukiDoll
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh woops

  5. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i thought it said 'any'

  6. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do you get the question now?

  7. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes

  8. UsukiDoll
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  13. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    notice 3 dark 17 non-dark is impossible

  14. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  16. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can i get a yes?

  18. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sure, i guess

  19. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lol

  20. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the explanation seems reasonable enough

  21. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    btw, what level of maths is this?

  26. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    not sure

  27. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i meant like are you still in school or university?

  28. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  29. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    im in university

  30. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  31. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i find this problem confusing

  32. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whoops, should have told you that at the start!

  33. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    im trying to force these into mutually exclusive possibilities

  34. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but are these equally likely?

  36. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  37. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  38. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  39. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  40. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what was your approach?

  41. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  42. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can you show me your work

  43. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i need your help :)

  44. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1366635152002:dw|

  45. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm..i guess you could use combinatorics

  46. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    does order matter?

  47. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  48. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  49. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what do you get from that?

  50. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  51. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i could have used letters,

  54. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sure

  55. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  59. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it's close to 4/5 at least

  61. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    perl your answer is right thx

  63. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and you could solve it using the complement approach .

  64. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    kabric, how do you know?

  65. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea, basically what i was thinking

  66. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    chelsea, how did you get nCr using pretty print?

  67. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i just typed it?

  68. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  69. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that is 8C8 , 16 C 12

  70. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    do you how it comes out the same ?

  71. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what do you mean?

  72. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you sort of lost me there

  73. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh ok sorry

  74. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1366637463112:dw|

  75. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes...

  76. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  78. Chelsea04
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sure

  79. perl
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  80. WalterDe
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Correct answer => http://629a3486.linkbucks.com

  81. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.