a) Probability that Mae wins 3 of the games.
That means Mae wins three games and loose two games, so:
P = (4/5)³ x (1/5)² = 0.02048
b) Mae wins either 4 or 5 of the games.
The probability Mae wins 4 games (win four games and loose one):
P₁ = (4/5)⁴ x (1/5) = 0.08192.
The probability Mae wins 5 games:
P₂ = (4/5)⁵ = 0.32768.
So, the desired probability is
P = P₁+P₂ = 0.08192 + 0.32768 = 0.4096.