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anonymous
 4 years ago
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
anonymous
 4 years ago
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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the chance a child gets a game is more than one, all the fractions are less than one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Bdude999 , probability is always less than or equal to 1 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@satellite73 , help please :)

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, not completely sure how to do this with permutations and combinations. I could do a listing, but that would be quite unefficient.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem. I have also faced difficulties in this problem :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But I warn that it would be too hectic. Permutation and Combination would be the right way

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

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

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0Actually, find the probability, sorry.

anonymous
 4 years ago
Best 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??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks :) Let me keep trying myself :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre64 , any help please ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0once 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
 4 years ago
Best ResponseYou've already chosen the best response.0It 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
 4 years ago
Best ResponseYou've already chosen the best response.0or can it be that 1 child gets all 5 games >not possible

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Since it is given in the question that we have to find the probability that each child should atleast get one game.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes 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
 4 years ago
Best ResponseYou've already chosen the best response.0well, 1112 1121 1211 2111 is your favoured cases and i we bruting out the total cases

anonymous
 4 years ago
Best 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here 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??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@shivam_bhalla Read about Stirling number of second kind

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.05000 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
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Probably this http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html I haven't (yet) encountered with anything better than these two.

anonymous
 4 years ago
Best 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 ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolForMath , Thanks bro, I got it :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If this question was diving 5 common things among four different children, then??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Read about "Stars and bars" combinatorics

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Will read about Stars and bars" combinatorics and report back here :)

anonymous
 4 years ago
Best 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 ??

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0with the atleast one constraint it will be the first one.
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