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?
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 $28\cdot\frac12=14$.