Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner.
A) Suppose that one of the teams is stronger than the other and wins each game with probability .6 independently of the outcomes of the other games. Find that probability that the stronger team wins the series. (Hint: let X be the random variable that denotes the number of games necessary until the stronger team wins 4.)

This implies that Probability that the stronger team wins the first 4 games or Probability that the stronger team wins 3 games in the first 4 games and Probability that the stronger team wins the fifth game or Probability that the stronger team wins 3 of the first 5 games and Probability that the stronger team wins the 6th game or Probability that the stronger team wins 3 games in the first 6 games.
This is the likely possibilities for the probability that the stronger team wins the series.
Applying the concept of combinatorics and binomial probability(P(x) = r) ; nCr x P^r x q^n-r
where q = probability of failure, probability of not winning = 1 - 0.4 = 0.6
= 0.6^4 + 4C1 x 0.6^4 x 0.4 + 5C2 x 0.6^4 x 0.4^2 + 6C3 x 0.6^4 x 0.4^3 = 0.710208
Therefore, probability that the stronger team wins the series = 0.710208