1. Let X = # of students who receive their own exam
(i.e., # of fixed points). Then E[X] = 1 and Var(X) = 1.
By Chebyshev's inequality,
Pr[|X-1| >= 4] <= Var(X)/4^2 = 1/16.
Now
Pr[X >= 5] <= Pr[|X-1| >= 4] <= 1/16,
since if X >= 5, then surely |X-1| >= 4.
Caution: We need to calculate Pr[|X-1| >= 4], not Pr[|X-1| >= 5].
The latter is equivalent to Pr[X >= 6], which is not what we were
asked to calculate. (The equivalence is because, in this case,
X is always non-negative. In general we would have Pr[X >= 5] <=
Pr[|X-1| >= 4] and Pr[X >= 6] <= Pr[|X-1| >= 5].)
2. Yes, this set is countable, since it is a subset of the natural
numbers. Therefore it can be put into a bijection with some subset
of N (just look at the identity function).
Alternatively, let S = the set of prime numbers. Define the function
f:N->S by f(n) = the nth prime number. For instance, f(0) = 2, f(1) = 3,
f(2) = 5, f(3) = 7, f(4) = 11, and so on. Then f is bijective, so S must
be countable.