Variance and applying Chebyshev's Inequality.

Question

Suppose that a certain conference has a random number of attendees. The expected number of attendees is 100. Every pair of people at the conference are introduced to each other, and the expected number of introduction is 4962.

(a) What is the expectation of the square of the number of attendees?

(b) What is the variance of the number of attendees?

(c) A year is a bad year if less than 90 people show up. Prove that the probability of a bad year is atmost 1/4

I assumed that there is N number of people. Then,

E(N) = 10

E(binomial(n, 2)) = 4962

I don't know how to relate E(N) to E(# of introduction).

Answer 1

Hints: $X=$ number of appearing attendees, $E(X)=100$. Then why is $E\bigl(\frac 12X(X-1)\bigr)=4962$? From here $E(X^2)=10024$, hence $V(X)=24$. Nice problem.