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­If 4 toys are randomly distributed among 4 children, what is the

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MartyMurray
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­If 4 toys are randomly distributed among 4 children, what is the [#permalink] New post   29 Feb 2024, 19:14
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­If 4 toys are randomly distributed among 4 children, what is the probability that no child will get more than two toys?

    (A) 1/16
     
    (B) 13/64
     
    (C) 13/16

    (D) 51/64
     
    (E) 15/16


Source: Marty Murray Coaching­­­­­­
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gmatophobia
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­If 4 toys are randomly distributed among 4 children, what is the [#permalink] New post   01 Mar 2024, 00:40
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MartyMurray wrote:
­If 4 toys are randomly distributed among 4 children, what is the probability that no child will get more than two toys?

    (A) 1/16
     
    (B) 13/64
     
    (C) 13/16

    (D) 51/64
     
    (E) 15/16


Source: Marty Murray Coaching­­­­­­


Let's find the opposite of what's asked i.e. the number of ways when a child can get more than two toys.

Case 1: 3 + 1 + 0 + 0

The four toys are distributed in a manner such that out of four children, one child gets three toys, another child gets one toy and the remaining two children get no toys.
  • Number of ways of selecting two children out of four children = \(^4C_2 = 6\) 
  • Number of ways of grouping the toys into bundles of 3 toys and 1 toy= \(^4C_3 * ^1C_1 = 4\) (i.e. Out of four toys, select three toys. From the remaining one toy, select one)
  • Number of ways of distributing the grouped toys = 2 (i.e. either the first child gets three toys and the second child gets one, or vice versa)

Total = \(48\)

Case 2: 4 + 0 + 0 + 0
­
The four toys are distributed in a manner such that out of four children, one child gets all four toys, and the other three children get none.
  • Number of ways of selecting one child out of four children = \(^4C_1 = 4\)
  • Number of ways of grouping the toys = \(^4C_4 = 1\) (i.e. Out of four toys, select all)
  • Number of ways of distributing the grouped toys = 1 (i.e. the child selected in the first step gets all toys)

Total = \(4 * 1 * 1 = 4\)

Required Probability

  • The number of ways when a child can get more than two toys = \(48 + 4 = 52\)
  • The number of ways of randomly distributing 4 toys among 4 children = \(4^4 = 256\)
  • The number of ways of randomly distributing 4 toys among 4 children such that no child will get more than two toys = 256 - 52 = 204

Required Probability = \(\frac{\text{Favourable Ways}}{\text{Total Number of ways}} = \frac{204}{256} = \frac{51}{64}\)

Option D



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DrAnkita91
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Re: ­If 4 toys are randomly distributed among 4 children, what is the [#permalink] New post   09 Mar 2024, 23:32
why is the formula (n+r-1 _C_ r-1) not working here?
4 toys are randomly distributed among 4 children = 7 _C_ 3 ways right?
So denominator should be this?
please correct me
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Re: ­If 4 toys are randomly distributed among 4 children, what is the [#permalink] New post   09 Mar 2024, 23:44
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DrAnkita91 wrote:
why is the formula (n+r-1 _C_ r-1) not working here?
4 toys are randomly distributed among 4 children = 7 _C_ 3 ways right?
So denominator should be this?
please correct me

­DrAnkita91 - This formula works when the toys are identical. The question doesn't mention that. 
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Re: ­If 4 toys are randomly distributed among 4 children, what is the [#permalink] New post   11 Mar 2024, 00:18
Antithesis:
Scenario when 1 kid get > 2 toys:

I. 1 kid get 3 toys and the other one get the rest

P1 = 4C1kid x 4C3 toy x 3C1 kid x 1C1 toy
= 4 x 4 x 3 x 1 = 3 x 4^2

II. 1 kid get 4 toys

P2 = 4C1 kid x 4C4 toy
= 4 x 1 = 4

Total P Ant= 3x4^2 + 4 = 4 (3x4 + 1) = 4 (13)

Total ways of distributing toys = 4 toys to 4 kids = 4^4

Probability of 1 kid get =<2 : 1- Total P Ant / Total Ways

Prob = 1 - {4 x 13 / 4^4)
= 1 - 13/64
= 51/64

ANS D
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Re: ­If 4 toys are randomly distributed among 4 children, what is the [#permalink]
11 Mar 2024, 00:18
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