Mainly e and f needed! 7. The Cubs play the Cardinals in the National League Cha

Question

Mainly e and f needed!

7. The Cubs play the Cardinals in the National League Championship Series in a best-of-seven series, i.e. a series which terminates when one of the two teams wins for the fourth time. Suppose that in each game played, the Cubs win with probability p independent of the outcomes of the other games. (a) The Cubs win in 7 games if an only if they win exactly 3 of the first 6 games, and then win the seventh game. Use this reasoning to find, as a function of p, the probability that the Cubs win the series in 7 games. (b) Use the same reasoning to find, as a function of p, the probability that the Cubs win in 6 games; in 5 games; and in 4 games. (c) Find the probability that the Cubs win the series in terms of p by summing the probabilities from (a) and (b). d) Find the probability that the Cubs win any 4 of the 7 games in terms of p. That, is assume that the Cubs and the Cardinals play all 7 games, no matter what. How does this compare to your answer from (c)? (e) Returning to the best-of-7 scenario (that is, the series stops once a team has won 4 games), let random variable N be the number of games played in the series. Find the probability mass function for N (you will need to find the probability that the Cardinals win in n games!) (f) What happens to the pmf as p approaches 1? What happens to the pmf as p approaches 0?

Explanation / Answer

Part (e)

Since it is best of 7, the minimum number of games to be played is 4 and it can be at most be 7. Thus, n = 4, 5, 6 or 7.

Now, n = 4 => the team all the first 4 games probability for which is p4

When n = 5, the team wins 3 out of first 4 games and then wins the fifth game.

Probability = (4C3)p3(1- p)p = 4p4(1- p)

When n = 6, the team wins 3 out of first 5 games and then wins the sixth game.

Probability = (5C3)p3(1- p)2p = 10p4(1- p)2

When n = 7, the team wins 3 out of first 6 games and then wins the seventh game.

Probability = (6C3)p3(1- p)3p = 20p4(1- p)3

Thus, probability mass function of N is:

n

4

5

6

7

p(n)

p4

4p4(1- p)

10p4(1- p)2

20p4(1- p)3

Or , in analytical form, p(n) = (n - 1)C3p4(1 - p)n – 4, n = 4, 5, 6, 7 ANSWER

Part (f)

As p approaches 1, (1 - p) approaches 0 and consequently, p(4) > p(5) > p(6) > p(7).

The distribution becomes heavily left skewed.

As p approaches 0, (1 - p) approaches 1 and consequently, p(4) < p(5) < p(6) < p(7).

The distribution becomes heavily right skewed.

DONE

n

4

5

6

7

p(n)

p4

4p4(1- p)

10p4(1- p)2

20p4(1- p)3

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