Hi, i hope u understand it with this proof:
• Let n be the pumping lemma value and let k be a prime
greater than n.
• If L is regular, the Pumping Lemma implies that a^k can be decomposed into
xyz, |y| > 0, such that xy^iz is in L for all i ≥ 0.
• Assume such a decomposition exists.
• The length of w = xy^k+1z must be a prime if w is in L. But

length(xyk+1z) = length(xyzyk)
= length(xyz) + length(yk)
= k + k(length(y)
= k (1 + length(y))
• The length of xy^k+1z is therefore not prime, since it is the
product of two numbers other than 1. So xy^k+1z is not in L.
• Contradiction!

I have this problem which says, "A programmer can use the random feature to emulate the flipping of a coin. For example, if the generator produces a number 0 or 1, the value 0 can be assigned heads and the value 1 can be assigned tails.
Write a program that ...