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# Mr. & Mrs Paul have 'a' number of children. The children wanted

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Manager
Joined: 08 Apr 2017
Posts: 68
Mr. & Mrs Paul have 'a' number of children. The children wanted  [#permalink]

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06 Jun 2018, 17:01
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Difficulty:

95% (hard)

Question Stats:

36% (01:58) correct 64% (01:56) wrong based on 87 sessions

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Mr. & Mrs Paul have 'a' number of children. The children wanted to gift something to their parents on their marriage anniversary. They shortlisted 'b' types of gifts and each child bought one gift for the parents. If each child thought of buying a different gift, what is the probability that, even after trying their best to buy different gifts, all of them bought the same gift, given b > a?

A. $$\frac{1}{b}^{a}$$

B. $$\frac{1}{b}^{(a-1)}$$

C. $$\frac{a}{b^{a}}$$

D. 1 - $$\frac{1}{b}^{(a-1)}$$

E. 1 - $$\frac{1}{b}^{a}$$
##### Most Helpful Expert Reply
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3367
Re: Mr. & Mrs Paul have 'a' number of children. The children wanted  [#permalink]

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07 Jun 2018, 01:00
2
4

Solution

Given:
• Mr. & Mrs. Paul have ‘a’ number of children
• The children shortlisted ‘b’ types of gifts and each of them bought one gift
• Also, $$b > a$$

To find:
• The probability that all of the children bought the same gift

Approach and Working:
As every child of the ‘a’ children can choose one type from ‘b’ types of gifts,
• The total number of ways they can choose gifts = $$b^a$$

Out of all the possible ways of choosing gifts, in only one case they will choose the same gifts

As total ‘b’ types of gifts are available,
• Number of ways they can choose the same gift = b

Therefore, the required probability = $$\frac{b}{b^a} = b^{(1-a)} = \frac{1}{b}^{(a-1)}$$

Hence, the correct answer is option B.

Answer: B
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##### General Discussion
Manager
Joined: 30 Jun 2019
Posts: 209
Mr. & Mrs Paul have 'a' number of children. The children wanted  [#permalink]

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14 Mar 2020, 16:16
Why is it equal to b^a and not simply b*a?

If i have 3 people and 4 catagories, my total is 3*4 combinations.
a1b1
a1b2
a1b3
a1b4
...
a3b3
a3b4
totaling to 12.

4^3 doesn't equal 12.

It looks like you're treating this as one co-dependent action not separate independent actions, and i'm not sure why.
The actions of one person picking a gift type does not influence the action of the next person picking a gift type, so the combinations should be independent and should be added, not multiplied.

Please explain why this situation is 4C1*4C1*4C1 and not 4C1+4C1+4C1.

Edit:
Nevermind - i understand my thought process mistake.
Adding would give you the selection of 1 kid OR another kid selecting a gift, not 1 kid AND another kid selecting a gift
a1b1 a2b1
a1b1 a2b2

they are actually codependent.
Intern
Joined: 14 Mar 2020
Posts: 3
Re: Mr. & Mrs Paul have 'a' number of children. The children wanted  [#permalink]

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14 Mar 2020, 19:40
Why isn’t the total number of ways a to the power of b rather then the other way?

Posted from my mobile device
Intern
Joined: 30 Sep 2019
Posts: 29
Re: Mr. & Mrs Paul have 'a' number of children. The children wanted  [#permalink]

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15 Mar 2020, 01:12
jacelynnp wrote:
Why isn’t the total number of ways a to the power of b rather then the other way?

Posted from my mobile device

Say a = 3 and b = 10.
The first kid can choose from 10 gifts, the second can choose from 10, the third can choose from 10.
There are 10 choices being made thrice; 10*10*10 = 10^(3) -> b^(a).
Re: Mr. & Mrs Paul have 'a' number of children. The children wanted   [#permalink] 15 Mar 2020, 01:12

# Mr. & Mrs Paul have 'a' number of children. The children wanted

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