Teams A and B play a series of games with the first team to win 4 games being declared the winner of the series. Suppose that team A independently wins each game with probability 0.6. Find the probability that team A wins the series.
P = .6^4 = .6 * .6 *.6 *.6 = ?
Assuming that A wins the first four games.
P = .6^4 = .6 * .6 *.6 *.6 = ?
. Assume you have applied to two different jobs A and B. In the past, 20 % of applicants who applied for Job A were offered jobs, while Job B offered 10 % of the applicants. Assume events are independent of each other.
(a) What is the probability that you will be offered both jobs? (b) What is the probability that you will be offered at least one job? (c) What is the probability that one and only one of the jobs will be offered to you? (d) What is the probability that neither jobs will be offered to you?
5
To find the probability that team A wins the series, we need to consider the different possible outcomes of the games.
Team A can win the series in the following ways:
- Winning all 4 games: This can happen with a probability of (0.6)^4 since each game is independent.
- Winning 3 games and losing 1: This can happen in 4 different ways (winning the first 3 games and losing the last, winning the first 2 games and losing the third, etc.). The probability of winning 3 games and losing 1 is (0.6)^3 * (1-0.6) = (0.6)^3 * (0.4).
- Winning 3 games and losing 2: This can happen in 10 different ways (choosing 2 out of the 5 games to lose). The probability of winning 3 games and losing 2 is (0.6)^3 * (0.4)^2.
- Winning 3 games and losing 3: This can happen in 10 different ways. The probability of winning 3 games and losing 3 is (0.6)^3 * (0.4)^3.
Now, we can calculate the total probability of team A winning the series by summing up all these possibilities:
P(A wins the series) = (0.6)^4 + 4 * (0.6)^3 * (0.4) + 10 * (0.6)^3 * (0.4)^2 + 10 * (0.6)^3 * (0.4)^3
Evaluating this expression will give the probability that team A wins the series.