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Suppose the probability of success obtaining heads is $0.5$, number of trials $n=10$. In a class of $100$ students, each student tosses a coin (assumed fair) $10$ times. Write down the expected number of students who will obtain $0,1,\ldots , 10$ heads.

If I start with $0$ heads, I understand that I need to calculate the probability of $0$ students obtaining $0$ heads, then the probability of $1$ student obtaining $0$ heads and so on until $100$. Then multiply the probability by the number of students and add all $100$ cases to get the expected number. Now obviously this is a lot of work, and this is just a part of a question, so is there something I don't understand or know, or is it the question?

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  • $\begingroup$ Can you calculate the probability that the first student tosses $k$ heads in $10$ tosses? How, if at all, does it differ from the probability that the second student tosses $k$ heads in $10$ tosses? $\endgroup$ – Dilip Sarwate Jan 17 '13 at 20:40
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Fortunately you don't need to do all that work. You can put to use the linearity of expectation. Denote the probability that a single student obtains $k$ heads by $p_k$. Then the expected number of students that this student contributes to the total number of students who obtain $k$ heads is $p_k\cdot1+(1-p_k)\cdot0=p_k$. Since the total number of students who obtain $k$ heads is the sum of $100$ contributions from the $100$ students, with each contribution given by $p_k$, by linearity of expectation the expected number of students who obtain $k$ heads is $100p_k$.

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What's the probability that a student gets $0$ head? Suppose it's $A$.

Then the expected number of students who get $0$ head would be $100\times A$ since the probability of getting $0$ head is the same for all students.

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Let $X$ = Number of heads.
$X$ ~ $B(10, .5)$ (that is, $X$ is a binomially distributed random variable), so the expected number of students who obtain $0,1,2,...10$ heads is as follows:

$100\cdot\Pr(X = 0) = 100\cdot{10\choose 0}0.5^0 (1-0.5)^{10-0}$
$100\cdot\Pr(X = 1) = 100\cdot{10\choose 1}0.5^1 (1-0.5)^{10-1}$
$100\cdot\Pr(X = 2) = 100\cdot{10\choose 2}0.5^2 (1-0.5)^{10-2}$
$100\cdot\Pr(X = 3) = 100\cdot{10\choose 3}0.5^3 (1-0.5)^{10-3}$

$100\cdot\Pr(X = 4) = 100\cdot{10\choose 4}0.5^4 (1-0.5)^{10-4}$
$100\cdot\Pr(X = 5) = 100\cdot{10\choose 5}0.5^5 (1-0.5)^{10-5}$
$100\cdot\Pr(X = 6) = 100\cdot{10\choose 6}0.5^6 (1-0.5)^{10-6}$
$100\cdot\Pr(X = 7) = 100\cdot{10\choose 7}0.5^7 (1-0.5)^{10-7}$

$100\cdot\Pr(X = 8) = 100\cdot{10\choose 8}0.5^8 (1-0.5)^{10-8}$
$100\cdot\Pr(X = 9) = 100\cdot{10\choose 9}0.5^9 (1-0.5)^{10-9}$
$100\cdot\Pr(X = 10) = 100\cdot{10\choose 10}0.5^{10} (1-0.5)^{10-10}$
This is due to the fact that:
$$ E[100\cdot X] = 100\cdot E[X] $$ and that $E[X=j] = 1\cdot P[X=j] + 0\cdot (1-P[X=j]) = P[X=j]$ for all $j \in \{0,1,...,10\}$

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