anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
none of these
anonymous
  • anonymous
the chance a child gets a game is more than one, all the fractions are less than one
anonymous
  • anonymous
@Bdude999 , probability is always less than or equal to 1 :)

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anonymous
  • anonymous
@satellite73 , help please :)
anonymous
  • anonymous
@precal , any idea??
inkyvoyd
  • inkyvoyd
Lemme look.
inkyvoyd
  • inkyvoyd
Sorry, not completely sure how to do this with permutations and combinations. I could do a listing, but that would be quite unefficient.
anonymous
  • anonymous
no problem. I have also faced difficulties in this problem :)
inkyvoyd
  • inkyvoyd
So, I can do listing?
anonymous
  • anonymous
Go ahead
anonymous
  • anonymous
But I warn that it would be too hectic. Permutation and Combination would be the right way
inkyvoyd
  • inkyvoyd
Ok, I think I have an idea.
inkyvoyd
  • inkyvoyd
First find the probability that the 4 games would be distributed evenly.
inkyvoyd
  • inkyvoyd
4!
inkyvoyd
  • inkyvoyd
Now, find the probability that ou of the 4 games, one got repeated.
inkyvoyd
  • inkyvoyd
*possibilities
inkyvoyd
  • inkyvoyd
Actually, find the probability, sorry.
anonymous
  • anonymous
I am trying..
precal
  • precal
sorry out of my league
anonymous
  • anonymous
@precal , out of my league too. Do you know who can solve this or who is master of permutations and combinations here??
precal
  • precal
sorry not at the moment
anonymous
  • anonymous
Thanks :) Let me keep trying myself :)
anonymous
  • anonymous
@amistre64 , any help please ?
amistre64
  • amistre64
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
or can it be that 1 child gets all 5 games --->not possible
anonymous
  • anonymous
It can be that 1 child gets 2 toys and rest of them get one each
amistre64
  • amistre64
if the toys are handed out at random, then i dont see why 1 child couldnt get all the toys
anonymous
  • anonymous
Since it is given in the question that we have to find the probability that each child should atleast get one game.
amistre64
  • amistre64
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?
amistre64
  • amistre64
otherwise there has to be the possibility that one child gets all the toys and such
anonymous
  • anonymous
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
amistre64
  • amistre64
well, 1112 1121 1211 2111 is your favoured cases and i we bruting out the total cases
anonymous
  • anonymous
@amistre64 , I have found a pre-written solution to this problem. But I am not able to understand the solution
anonymous
  • anonymous
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??
anonymous
  • anonymous
@shivam_bhalla Read about Stirling number of second kind
amistre64
  • amistre64
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
@FoolForMath , can you suggest a better source because I am totally new to this. Thanks :) @amistre64 ,Thanks for helping a lot mate :)
anonymous
  • anonymous
Probably this http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html I haven't (yet) encountered with anything better than these two.
anonymous
  • anonymous
@FoolForMath , thanks a lot mate :) You saved my day :) If I have any doubts regarding this, can I message you ??
anonymous
  • anonymous
@FoolForMath , Thanks bro, I got it :)
anonymous
  • anonymous
Glad to help.
anonymous
  • anonymous
If this question was diving 5 common things among four different children, then??
anonymous
  • anonymous
Read about "Stars and bars" combinatorics
anonymous
  • anonymous
Will read about Stars and bars" combinatorics and report back here :)
anonymous
  • anonymous
@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 ??
anonymous
  • anonymous
@FoolForMath Or the answer is \[(5+4-1) C _{4-1}= 8 C _{3}=56\]
anonymous
  • anonymous
with the at-least one constraint it will be the first one.

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