Suppose that a child desires 10 different toys for her birthday. Twenty people will come to her birthday party, each of them equally likely to bring any one of the 10 toys. Let X be the number of different types of toys brought to the party. Note that X can be any integer from 1 to 10. What is E[X]? Justify your answer.

8.784

Step-by-step explanation:

There are 10 different toys, hence let

i = 1, 2, ... 10

The expected value for each different toys is debited such that E (Xi). For example, the expected value for the first toy is E (X1)

Since 10 toys, so we have E (X1), E (X2), ... (E (X10)

Total expectation value E (X) will be

E (X) = E (X1) + E (X2) + ... (E (X10)

Then, to make it simpler, we set the condition of the probability by setting the value of getting a toy to be 1 and not getting a toy to be 0

Also note that the number of people coming to the party is 20 and they are equally likely to bring any one of the 10 toys.

Therefore,

E (Xi) = P (Xi=1) = 1 - (9/10) ^20

So for total expectation value,

E (X) = E (X1) + E (X2) + ... (E (X10)

= 10 * (1 - (9/10) ^20)

= 8.784