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Hi I was working on this question for my exam review:

Consider a multiple choice exam with 100 questions, in which for each question, four options are given to choose from. You answer each question by choosing an answer uniformly at random, and independently of the other answers. What is the expected number of correct answers?

After doing some simple calulation I ended up with 25 correct answers however the textbook answers page says $\sum\limits_{k=0}^{100}k {100 \choose k} (1/4)^k (3/4)^{100-k}$ I have no idea how it ended up with that answer any help?

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  • $\begingroup$ 25 is correct. The expression comes from the the formula for expectation and binomial distribution( en.wikipedia.org/wiki/Binomial_distribution ) which closely follows Binomial formula. The expression given on the last line evaluates to 25. $\endgroup$ – talegari Apr 25 '14 at 19:23
  • $\begingroup$ I just rechecked it does multiply by k anyways how would I do this? $\endgroup$ – alli Apr 25 '14 at 19:23
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The expected number of correct solutions is the sum of all posible scores mutiplied by the probability of getting each of those scores. What is the probability you get $k$ answers?

you need to get $k$ correct and $100-k$ incorrect. so assuming you had to get the first k correct and the last 100-k incorrect the probability would be $(\frac{1}{4})^k(\frac{3}{4})^{100-k}$. However there are $\binom{100}{k}$ different ways to pick the $k$ answers you got right. Thus the probability you get $k$ points is:

$\binom{100}{k}(\frac{1}{4})^k(\frac{3}{4})^{100-k}$

And we know the estimated number of answers is sum of each value multiplied by its probability, so we get:

$\sum_{k=0}^{100}k\binom{100}{k}(\frac{1}{4})^k(\frac{3}{4})^{100-k}$

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They are computing the probability of $k$ correct answers as chosing $k$ correct ones out of $100$ (and $100 - k$ failed), and then the respective probabilities of success/failure.

Also note that: \begin{align} \sum_{k} k \binom{n}{k} z^k (1 - p)^{n - k} &= z \frac{\mathrm{d}}{\mathrm{d} z} \sum_k \binom{n}{k} z^k (1 - p)^{n - k} \\ &= z \frac{\mathrm{d}}{\mathrm{d} z} (1 - p + z)^n \\ &= n z (1 - p + z)^{n - 1} \end{align} If you evaluate the last sum at $z = p$ you have: $$ \sum_{k} k \binom{n}{k} p^k (1 - p)^{n - k} = n p $$ Just as you obtained.

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