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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 $E[X]$?

I am asked to calculate $E[x]$.

I am basically computing $$E[X]E[X] = (1)P(1 \text{ toy type}) + (2)P(2 \text{ toy types}) + (3)P(3 \text{ toy types}) + (4)P(4 \text{ toy types}) + (5)P(5 \text{ toy types}) + \ldots$$ all the way to $10$ toy types.

I don't know if its right. Please give me a short/brief solution that I can expand on myself.

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3 Answers 3

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A short/brief hint that you can expand on yourself:

The easiest way to do this is to calculate the probability a particular toy is or is not brought and then use linearity of expectation

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Let $X_i$ be random variable which is $1$ if child gets present $i$ and $0$ if not.

Then $X= X_1+...+X_{10}$ and $$E(X_i) = P(X_i=1) = 1-\Big({9\over 10}\Big)^{20}$$

So $$E(X) = E(X_1)+...+E(X_{10}) = 10\Big[ 1-\Big({9\over 10}\Big)^{20}\Big]$$

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  • $\begingroup$ why does E(X)=E(X1)+...+E(X10)? i don't understand why expected value of number of toys equal of the sum of probability of each toy being brought to the party. thanks $\endgroup$ Commented Dec 4, 2017 at 4:27
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[This is the birthday problem with 10 days per year and 20 persons.] The probability that $t$ toys are missing is for $t=0..10$ in a table of $t$, $p(t)$ as an exact rational number and the floating point fraction:

0 671054134991877/3125000000000000 0.21473732319740063
1 272415890382363/625000000000000 0.4358654246117808
2 8601918819738993/31250000000000000 0.2752614022316478
3 526532928657057/7812500000000000 0.0673962148681033
4 101731113903447/15625000000000000 0.006510791289820608
5 283199902209/1250000000000000 2.265599217672E-4
6 2849623301763/1250000000000000000 2.2796986414104E-6
7 2612729007/625000000000000000 4.1803664112E-9
8 4718583/10000000000000000000 4.718583E-13
9 1/10000000000000000000 1.0E-19
10 0 

$$ \sum_{t=0}^{10} p(t)=1. $$ The probability that $t$ toys are received is $p(10-t)$ and the expectation value is $\sum_t tp(10-t)=8.784233454$, confirming the @nonuser anwer.

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