Question 850137
Determine the probability of Team A winning 0 or 1 game, then we can subtract from 1.

p = 0.45 (Team A wins, B loses)
q = 0.55 (Team B wins, A loses)

For 0 wins, that would be the probability of 4 wins by team B.
P(x=0) = q^4
= (0.55)^4
= 0.55 * 0.55 * 0.55 * 0.55
= 0.09150625
≈ 9.15%

For 1 win, that would be the probability of 1 win by A and 3 wins by B.
You might think that would just be 0.45 * 0.55 * 0.55 * 0.55, but there are 4 ways to pick which of the matches is won by A.
P(x=1) = C(4,1) * p^1 * q^3
= 4 * (0.45) * (0.55)^3
= 4 * 0.45 * 0.55 * 0.55 * 0.55
= 0.299475
≈ 29.95%

P(x=0 or 1) = P(x=0) + P(x=1)
= 0.0915 + 0.2995
= 0.3910
= 39.1%

But you want the probability of 2 or more which is the opposite probability:
P(x ≥ 2) = 1 - P(x = 0 or 1)
= 1 - 0.3910
= 0.6090
= 60.9%

Answer:
P(x ≥ 2) = 60.9%