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shivam_bhalla Group Title

5 different games are to be distributes among 4 children randomly. The probabilty that each child get altleast one game is A)1/4 B)15/64 c) 21/64 d) none of these

  • 2 years ago
  • 2 years ago

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  1. Bdude999 Group Title
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    none of these

    • 2 years ago
  2. Bdude999 Group Title
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    the chance a child gets a game is more than one, all the fractions are less than one

    • 2 years ago
  3. shivam_bhalla Group Title
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    @Bdude999 , probability is always less than or equal to 1 :)

    • 2 years ago
  4. shivam_bhalla Group Title
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    @satellite73 , help please :)

    • 2 years ago
  5. shivam_bhalla Group Title
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    @precal , any idea??

    • 2 years ago
  6. inkyvoyd Group Title
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    Lemme look.

    • 2 years ago
  7. inkyvoyd Group Title
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    Sorry, not completely sure how to do this with permutations and combinations. I could do a listing, but that would be quite unefficient.

    • 2 years ago
  8. shivam_bhalla Group Title
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    no problem. I have also faced difficulties in this problem :)

    • 2 years ago
  9. inkyvoyd Group Title
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    So, I can do listing?

    • 2 years ago
  10. shivam_bhalla Group Title
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    Go ahead

    • 2 years ago
  11. shivam_bhalla Group Title
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    But I warn that it would be too hectic. Permutation and Combination would be the right way

    • 2 years ago
  12. inkyvoyd Group Title
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    Ok, I think I have an idea.

    • 2 years ago
  13. inkyvoyd Group Title
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    First find the probability that the 4 games would be distributed evenly.

    • 2 years ago
  14. inkyvoyd Group Title
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    4!

    • 2 years ago
  15. inkyvoyd Group Title
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    Now, find the probability that ou of the 4 games, one got repeated.

    • 2 years ago
  16. inkyvoyd Group Title
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    *possibilities

    • 2 years ago
  17. inkyvoyd Group Title
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    Actually, find the probability, sorry.

    • 2 years ago
  18. shivam_bhalla Group Title
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    I am trying..

    • 2 years ago
  19. precal Group Title
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    sorry out of my league

    • 2 years ago
  20. shivam_bhalla Group Title
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    @precal , out of my league too. Do you know who can solve this or who is master of permutations and combinations here??

    • 2 years ago
  21. precal Group Title
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    sorry not at the moment

    • 2 years ago
  22. shivam_bhalla Group Title
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    Thanks :) Let me keep trying myself :)

    • 2 years ago
  23. shivam_bhalla Group Title
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    @amistre64 , any help please ?

    • 2 years ago
  24. amistre64 Group Title
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    once a child gets a game, are they out of rotation until they all have games? or can it be that 1 child gets all 5 games?

    • 2 years ago
  25. shivam_bhalla Group Title
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    It is like there are 5 different toys and 4 children and we have to find the probability that each child should get atleast one toy

    • 2 years ago
  26. shivam_bhalla Group Title
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    or can it be that 1 child gets all 5 games --->not possible

    • 2 years ago
  27. shivam_bhalla Group Title
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    It can be that 1 child gets 2 toys and rest of them get one each

    • 2 years ago
  28. amistre64 Group Title
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    if the toys are handed out at random, then i dont see why 1 child couldnt get all the toys

    • 2 years ago
  29. shivam_bhalla Group Title
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    Since it is given in the question that we have to find the probability that each child should atleast get one game.

    • 2 years ago
  30. amistre64 Group Title
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    if each child gets a toy and there is one left over; then the prob that each child gets at least 1 toy is: 1 right?

    • 2 years ago
  31. amistre64 Group Title
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    otherwise there has to be the possibility that one child gets all the toys and such

    • 2 years ago
  32. shivam_bhalla Group Title
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    Yes you are right. But the question is not that . It is what is the probability that each boy has atleast one toy. Which means like 21111 , 12111, 11211,11121,11112 --> These are the favourable cases(accd to question). Now the problem is I need to know total no of cases. to get my probability

    • 2 years ago
  33. amistre64 Group Title
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    well, 1112 1121 1211 2111 is your favoured cases and i we bruting out the total cases

    • 2 years ago
  34. shivam_bhalla Group Title
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    @amistre64 , I have found a pre-written solution to this problem. But I am not able to understand the solution

    • 2 years ago
  35. shivam_bhalla Group Title
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    Here is the solution n(S) = 4^5 Total ways of distribution so that each child gets atleast one game \[=4^{5}-4 C _{1} * 3^{5} + 4C_{2} * 2^{5}-4C _{3}\] = 1024- 4*243 + *32-4 = 240 Reqd probability = 240 / 4^5 = 15/64 Now I don't understand the total ways of distribution in the solution??

    • 2 years ago
  36. FoolForMath Group Title
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    @shivam_bhalla Read about Stirling number of second kind

    • 2 years ago
  37. amistre64 Group Title
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    5000 0500 0050 0005 : 4 4100 3200 4010 ... 4001 0410 0401 1400 0041 1040 0140 1004 0104 0014 : 24 0122 0113 0212 ... 0221 1022 1202 1220 2012 2021 2102 2120 2201 2210 : 24 1112 1121 1211 2111 :4 56 altogether? you see any i missed?

    • 2 years ago
  38. FoolForMath Group Title
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    The ways of distribution is consistent to dividing r distinct things into n distinct groups \( n! \times S(n,r) \). If you expand this you will find the closed form which is used in your solution. REF:http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Definition

    • 2 years ago
  39. shivam_bhalla Group Title
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    @FoolForMath , can you suggest a better source because I am totally new to this. Thanks :) @amistre64 ,Thanks for helping a lot mate :)

    • 2 years ago
  40. FoolForMath Group Title
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    Probably this http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html I haven't (yet) encountered with anything better than these two.

    • 2 years ago
  41. shivam_bhalla Group Title
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    @FoolForMath , thanks a lot mate :) You saved my day :) If I have any doubts regarding this, can I message you ??

    • 2 years ago
  42. shivam_bhalla Group Title
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    @FoolForMath , Thanks bro, I got it :)

    • 2 years ago
  43. FoolForMath Group Title
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    Glad to help.

    • 2 years ago
  44. shivam_bhalla Group Title
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    If this question was diving 5 common things among four different children, then??

    • 2 years ago
  45. FoolForMath Group Title
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    Read about "Stars and bars" combinatorics

    • 2 years ago
  46. shivam_bhalla Group Title
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    Will read about Stars and bars" combinatorics and report back here :)

    • 2 years ago
  47. shivam_bhalla Group Title
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    @FoolForMath if this question was dividing 5 common things among four different children, then the answer would be \[(5-1)C _{4-1}= 4C _{3} = 4\] ?? And what should be the total no of cases ??

    • 2 years ago
  48. shivam_bhalla Group Title
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    @FoolForMath Or the answer is \[(5+4-1) C _{4-1}= 8 C _{3}=56\]

    • 2 years ago
  49. FoolForMath Group Title
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    with the at-least one constraint it will be the first one.

    • 2 years ago
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