Question

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

Answer 1

none of these

Answer 2

the chance a child gets a game is more than one, all the fractions are less than one

Answer 3

@Bdude999 , probability is always less than or equal to 1 :)

Answer 4

@satellite73 , help please :)

Answer 5

@precal , any idea??

Answer 6

Lemme look.

Answer 7

Sorry, not completely sure how to do this with permutations and combinations. I could do a listing, but that would be quite unefficient.

Answer 8

no problem. I have also faced difficulties in this problem :)

Answer 9

So, I can do listing?

Answer 10

Go ahead

Answer 11

But I warn that it would be too hectic. Permutation and Combination would be the right way

Answer 12

Ok, I think I have an idea.

Answer 13

First find the probability that the 4 games would be distributed evenly.

Answer 14

4!

Answer 15

Now, find the probability that ou of the 4 games, one got repeated.

Answer 16

*possibilities

Answer 17

Actually, find the probability, sorry.

Answer 18

I am trying..

Answer 19

sorry out of my league

Answer 20

@precal , out of my league too. Do you know who can solve this or who is master of permutations and combinations here??

Answer 21

sorry not at the moment

Answer 22

Thanks :) Let me keep trying myself :)

Answer 23

@amistre64 , any help please ?

Answer 24

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?

Answer 25

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

Answer 26

or can it be that 1 child gets all 5 games --->not possible

Answer 27

It can be that 1 child gets 2 toys and rest of them get one each

Answer 28

if the toys are handed out at random, then i dont see why 1 child couldnt get all the toys

Answer 29

Since it is given in the question that we have to find the probability that each child should atleast get one game.

Answer 30

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?

Answer 31

otherwise there has to be the possibility that one child gets all the toys and such

Answer 32

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

Answer 33

well,

1112
1121
1211
2111 is your favoured cases

and i we bruting out the total cases

Answer 34

@amistre64 ,
I have found a pre-written solution to this problem. But I am not able to understand the solution

Answer 35

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??

Answer 36

@shivam_bhalla Read about Stirling number of second kind

Answer 37

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?

Answer 38

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

Answer 39

@FoolForMath , can you suggest a better source because I am totally new to this. Thanks :)
@amistre64 ,Thanks for helping a lot mate :)

Answer 40

Probably this http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html

I haven't (yet) encountered with anything better than these two.

Answer 41

@FoolForMath , thanks a lot mate :) You saved my day :) If I have any doubts regarding this, can I message you ??

Answer 42

@FoolForMath , Thanks bro, I got it :)

Answer 43

Glad to help.

Answer 44

If this question was diving 5 common things among four different children, then??

Answer 45

Read about "Stars and bars" combinatorics

Answer 46

Will read about Stars and bars" combinatorics and report back here :)

Answer 47

@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 ??

Answer 48

@FoolForMath Or the answer is
\[(5+4-1) C _{4-1}= 8 C _{3}=56\]

Answer 49

with the at-least one constraint it will be the first one.