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.)

Step-by-step explanation:

Considering for for stronger team, given that: P (winning) = 0.6 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