# Expected Number of Successes in a Sample

$200$ calculators are ordered and of those $200$, $20$ are broken. $10$ calculators are selected at random. Calculate the expected value of broken calculators in the selection.

Solution:

Chance of broken calculator: $\dfrac{1}{10}$.

Do I need to calculate the odds of $0$ - $10$ calculators being broken, multiply the probabilities with the respective $0$ - $10$ and add those together?

For example:

\begin{align} {10 \choose 0} \cdot \left(\frac{1}{10}\right)^0 \cdot \left(\frac{9}{10}\right)^{10} &= 0.3487 \tag{0}, \\ {10 \choose 1} \cdot \left(\frac{1}{10}\right)^1 \cdot \left(\frac{9}{10}\right)^9 &= 0.3874 \tag{1}, \\ \vdots & \\ {10 \choose 10} \cdot \left(\frac{1}{10}\right)^{10} \cdot \left(\frac{9}{10}\right)^0 &= 1 \cdot 10^{-10}. \tag{10} \end{align}

\begin{align} E[X] = (0) \cdot 0 + (1) \cdot 1 + \ldots + (10) \cdot 10. \end{align}

Or is it enough to calculate

\begin{align} E[X] &= (1+2+3+4+5+6+7+8+9+10) \cdot \left(\frac{1}{10}\right). \end{align}

Thanks.

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You're assuming sampling with replacement when you're calculating the probabilities. I'm not sure you can do that. If you can, then it's standard knowledge that the expected value of a binomial distribution with $n$ trials and probability $p$ of success is $np$, so in this case $E[X] = 200\cdot20/200 = 20$. – Michael Zhao Oct 21 '12 at 16:14
@MichaelZhao You have misread the problem. Only ten calculators are selected, and we are asked to find the expected number of broken calculators in that selection. – Austin Mohr Oct 21 '12 at 18:06
It is probably easier to compute the expectation directly. The answer by Alan below produces the same result but you need a little more work (Vandermonde's identity?) to compute the final result. – copper.hat Oct 21 '12 at 23:37
@Austin Mohr: Thanks. At any rate, it depends on whether you're sampling with replacement or without replacement (are the probabilities of being broken independent from one calculator to the next?). If you're sampling with replacement (probabilities are independent), then you still use the fact that if $X \sim B(n,p)$, then $E[X] = np = 10\cdot 20/200 = 1$, as below. – Michael Zhao Oct 22 '12 at 19:58

(See note below for the correct answer.)

Let $X_k = 1$ if the $k$th calculator is broken, and $0$ otherwise.

Assuming independence of 'brokenness', we have the probability of the $k$th calculator being broken is $P(X_k = 1) = \frac{20}{200} = \frac{1}{10}$.

Then the expected number of broken calculators is $$E (\sum_{k=1}^{10} X_k) = \sum_{k=1}^{10}1 P(X_k) = 10 \frac{1}{10} = 1$$

NOTE: The above relies on the randomness of selection, not of any characteristic of 'brokenness'. Hence the reasoning is incorrect, even if the answer is correct. Here is a correct answer with correct reasoning.

Let $J = \{1,...,200\}$ represent the calculators. Let $B \subset J$, with $|B| = 20$, represent the (indices of) the broken calculators. Let $n = 10$ represent the number of calculators selected.

Our sample space is $\Omega = \{I \subset J | |I| = n \}$. Assuming the sample is selected uniformly, we have the probability $P(I) = \frac{1}{|\Omega|}$ for all $I \in \Omega$.

As above, let $X_k = 1$ if the $k$th calculator is broken, and $0$ otherwise, that is $X_k = 1_{B}(k)$. Obviously, we have $\sum_{k \in J} X_k = |B|$. Let $I \in \Omega$, then $\sum_{k \in I} X_k$ represents the number of broken calculators in the sample $I$.

We wish to compute $E (\sum_{k \in I} X_k)$, and since the distribution is uniform, we have $E (\sum_{k \in I} X_k) = \sum_{I \in \Omega} \sum_{k \in I} X_k \frac{1}{|\Omega|}$.

To simplify this expression, count the number of times a given $X_k$ occurs in the above summation. There are a total of $n|\Omega|$ '$X_*$s, so any particular $X_k$ occurs $\frac{n|\Omega|}{|J|}$ times. Consequently we have

$$E (\sum_{k \in I} X_k) = \sum_{I \in \Omega} \sum_{k \in I} X_k \frac{1}{|\Omega|} = \frac{n|\Omega|}{|J|} \frac{1}{|\Omega|} \sum_{k \in J} X_k = \frac{n|B|}{|J|} = \frac{10\cdot 20}{200} = 1$$

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$\left({20 \choose 1}{180 \choose 9} + 2{20 \choose 2}{180 \choose 8} + \cdots + 10{20 \choose 10}{180 \choose 0} \right) / {200 \choose 10}$, where ${n \choose k}$ denotes the binomial coefficient ${n \choose k} = \frac{n!}{k!(n-k)!}$.

This follows because the probability $p_k$ that you your sample has $k$ broken calculators is ${20 \choose k}{180 \choose 10-k} / {200 \choose 10}$. Why? Because the denominator is counting the total number of ways you could sample $10$ calculators from a set of $200$, and the numerator is counting the total number of such subsets which have $k$ broken calculators, because it's equivalent to sampling $k$ calculators from the set of broken ones and $10-k$ calculators from the set of non-broken ones.

The expected value is therefor $p_1 + 2p_2 + 3p_3 + \cdots + 10p_{10}$ which is equal to the number I have on top.

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+1 If you compute the number, you will get $1$. – copper.hat Oct 21 '12 at 23:39

Your shorter formula doesn't work. Suppose the question asked you to select two calculators instead of ten. To compute the expected value "by hand", there are three cases to consider (we don't care about the case when no calculators are broken).

First Calculator Broken: This event happens with probability $(1/10)(9/10) = 9/100$.

Second Calculator Broken: This event happens with probability $(9/10)(1/10) = 9/100$.

Both Calculators Broken: This event happens with probability $(1/10)(1/10) = 1/100$.

The expected number of broken calculators (in my modified example) is therefore $$(1 \cdot 9/100) + (1 \cdot 9/100) + (2 \cdot 1/100) = 20/100 = 0.2.$$

The fact that we had two instances of "one calculator broken" would normally be accounted for by the factor of $C(2,1) = 2$.

To address your concern more directly, your shorter formula does not reflect the fact that "exactly one broken calculator" really means "one is broken and nine are functional", which is why that formula is giving a higher expected value than is correct.

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