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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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.

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).

Explanation:
GIVEN:

Number of children= a

Number of gifts shortlisted = b

Also, b > a

We have to find the probability that all the children will select the same gift.

Probability =FavorablenumberofoutcomesTotalnumberofoutcomes

Total number of outcomes:-

Total number of outcomes over here signifies the total number of different ways in which ‘a’

number of children can select 1 gift each out of ‘b’ gifts available.

Number of ways in which first child can select 1 gift out of ‘b’ gifts = bC1=b

Number of ways in which second child can select 1 gift out of ‘b’ gifts = bC1=b

Similarly, Number of ways in which every child can select 1 gift = b

So, Total number of ways = b×b×b×b×b…………a times =ba

Favorable number of outcomes:-

Favorable number of outcomes means the total number of different ways in which each child will

select the same gift out of ‘b’ gifts available.

So every child will either select first gift or second or third and so on.And as there are ‘b’ gifts

available, so there are ‘b’ options available to selectone gift .

So, Favorable number of outcomes = b

Therefore, Probability = b/(b)^a=(1/b)^(a−1)

Hence, the correct answer is B.

Probability that the first child chooses one of 'b' gifts is 1/b.
Probability that the second child chooses one of 'b' gifts is 1/b.
.
.
Probability that the 'a'th child chooses one of 'b' gifts is 1/b.
Probability that each child chooses any one of gift = 1/b * 1/b * ...... * 1/b (a times) = (1/b)^a
The no of ways of choosing 1 gift is b.
Hence, the probability that each child chooses same gift =[ b * (1/b)^a ] = b^(1-a).

Solution:

The first child can buy any gift as he or she chooses. The second child has a 1/b chance of buying the same gift as the first child. The third child also has a 1/b chance of buying the same gift as the first child. In other words, all the a - 1 children (other than the first child) have a 1/b chance of buying the same gift as the first child. Therefore, the probability of all the children buying the same give is (1/b)^(a - 1).

Answer: B
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Let a=1 and b=2, implying that ONE child shortlisted TWO gifts and that this child bought ONE of the two shortlisted gifts.
What is the probability that all of them bought the same gift?
Since there is only one child, the probability = 1.

The correct answer must yield 1 when a=1 and b=2.
Only B works:
\(\frac{1}{b}^{(a-1)} = \frac{1}{2}^0\) = 1


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