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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
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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Bdude999 Group TitleBest ResponseYou've already chosen the best response.0
none of these
 2 years ago

Bdude999 Group TitleBest ResponseYou've already chosen the best response.0
the chance a child gets a game is more than one, all the fractions are less than one
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@Bdude999 , probability is always less than or equal to 1 :)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 , help please :)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@precal , any idea??
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Lemme look.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
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

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
no problem. I have also faced difficulties in this problem :)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
So, I can do listing?
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
Go ahead
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
But I warn that it would be too hectic. Permutation and Combination would be the right way
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Ok, I think I have an idea.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
First find the probability that the 4 games would be distributed evenly.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Now, find the probability that ou of the 4 games, one got repeated.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
*possibilities
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Actually, find the probability, sorry.
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
I am trying..
 2 years ago

precal Group TitleBest ResponseYou've already chosen the best response.0
sorry out of my league
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@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

precal Group TitleBest ResponseYou've already chosen the best response.0
sorry not at the moment
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
Thanks :) Let me keep trying myself :)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 , any help please ?
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
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

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
or can it be that 1 child gets all 5 games >not possible
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
It can be that 1 child gets 2 toys and rest of them get one each
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
if the toys are handed out at random, then i dont see why 1 child couldnt get all the toys
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
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

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
otherwise there has to be the possibility that one child gets all the toys and such
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
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

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
well, 1112 1121 1211 2111 is your favoured cases and i we bruting out the total cases
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 , I have found a prewritten solution to this problem. But I am not able to understand the solution
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
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 + *324 = 240 Reqd probability = 240 / 4^5 = 15/64 Now I don't understand the total ways of distribution in the solution??
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
@shivam_bhalla Read about Stirling number of second kind
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
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

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@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

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Probably this http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html I haven't (yet) encountered with anything better than these two.
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@FoolForMath , thanks a lot mate :) You saved my day :) If I have any doubts regarding this, can I message you ??
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@FoolForMath , Thanks bro, I got it :)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Glad to help.
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
If this question was diving 5 common things among four different children, then??
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
Read about "Stars and bars" combinatorics
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
Will read about Stars and bars" combinatorics and report back here :)
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@FoolForMath if this question was dividing 5 common things among four different children, then the answer would be \[(51)C _{41}= 4C _{3} = 4\] ?? And what should be the total no of cases ??
 2 years ago

shivam_bhalla Group TitleBest ResponseYou've already chosen the best response.0
@FoolForMath Or the answer is \[(5+41) C _{41}= 8 C _{3}=56\]
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.2
with the atleast one constraint it will be the first one.
 2 years ago
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