G(V,E) is a simple graph with 8 vertices. The edges of G are decided by tossing the coin for each 2 vertices combination. Edge is added between any two vertices iff head is turned up. Expected number of edges in the Graph G(V,E) is.

I thought of doing like this

Let X be a random variable denoting the number of edges in the graph.

Total possible edges->$_8C_2=28$

Now, for each of those 28 edges, we tossed a coin and if it turned out to be heads,that edge was included.

(1)X=0,P(X)=$_{28}C_0 \frac{1}{2^{28}}$. All those 28 tosses of coin are independent with probability of heads=Probability of getting tails=$\frac{1}{2}$

(2)X=1(One edge), P(X)=$_{28}C_1 \times\frac{1}{2} \times \frac{1}{2^{27}}$-Means in those 28 trials, exactly 1 heads and rest tails.

(3)Similarly for all 28 edges-> $X=28,P(X)=_{28}C_{28} \times \frac{1}{2^{28}}$

And then we add all cases of X from 0 to 28 because $E[X]=\sum x.p(x)$

But this all together seems to be a very huge number and answer is given to be 14, where I am wrong in my reasoning?

  • $\begingroup$ Two remarks: (1) What you tried to write, is not very clear. You do not define what the $C_0,C_1,C_2,...$ are. (2) The problem becomes much simpler if you were to use the linearity of the expectation. By finding a collection of 'simple' random variables for which the calculation of expectation is simple, and that sum up to $X$, you will have simplified the problem many times over. $\endgroup$ – Keen-ameteur Dec 17 '18 at 6:31
  • $\begingroup$ It is not such a "very huge number" because the factor $\frac1{2^{28}}$ is pretty small. $\endgroup$ – bof Dec 17 '18 at 7:06

In other words, you want to know the expected number of times you're going to get heads in $28$ independent tosses of a fair coin. To do it your way, you can add up the $29$ terms by hand, or you can use the identity $$\sum_{k=0}^nk\binom nk=n2^{n-1}\tag1$$ which is obtained by differentiating the binomial identity $$(1+x)^n=\sum_{k=0}^n\binom nkx^n$$ with respect to $x$ and then setting $x=1$. Using $(1)$ we get $$E(X)=\sum_{k=0}^{28}k\binom{28}k\left(\frac12\right)^k\left(\frac12\right)^{28-k}=\left(\frac12\right)^{28}\sum_{k=0}^{28}k\binom{28}k=\left(\frac12\right)^{28}\cdot28\cdot2^{27}=14.$$

This shows that, if you toss a coin $28$ times, on average you're going to get $14$ heads. You can get the same result more easily by using the additivity of expectations: if $X$ is the sum of $28$ random variables, each of which has an expected value of $\frac12$, then the expected value of $X$ is $28\cdot\frac12=14$.

  • 1
    $\begingroup$ Another way to find $\sum_{k=0}^n k \binom nk$ is to substitute $\binom nk = \frac nk \binom{n-1}{k-1}$ in all except the $k=0$ term, getting $\sum_{k=1}^n n \binom{n-1}{k-1} = n \sum_{j=0}^{n-1}\binom{n-1}{j} = n2^{n-1}$. $\endgroup$ – Misha Lavrov Dec 18 '18 at 4:18

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