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

Why isn’t the total number of ways a to the power of b rather then the other way?

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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

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

Hey, why is the Number of ways they can choose the same gift = b ??

They each choose the the same gift, 1 gift.

If there are 3 gifts to be chosen among, there are 3 ways they can choose that one gift

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Aren't A and B choices the same thing? Because 1 squared, 1 cubed or something are always 1?

\(\frac{1}{b}^{a}\) => \(1^a\) / \(b^a\)

\(\frac{1}{b}^{(a-1)}\) => \(1^{a-1}\) / \(b^{a-1}\)

Depending on the value of b, both options will have different outcomes.

Strictly speaking the expressions should be bracketed to properly communicate that the exponent applies to both the numerator and denominator