quantskillsgmat wrote:Q baseball world series matches 2 teams against each other in best of seven series.The first team to win four games wins the series and no subsequent games are played.If you have no special information about either of two teams,what is probability that world series will consist of fewer than 7 games.
a)0.125 b)0.25 c)0.3125 d)0.6875 e)0.75
STEP 1: What are we counting? We are counting distinct world-series game arrangements.
STEP 2: What defines a world-series game arrangmeent?
A world-series game arrangement is a specific ordering of world-series games. Each game has two possible values: {A,B}, where A represents one team winning, and B represents the other team winning. The number of series games varies from 4 to 7 games, since the series ends when we find either 4 A's or 4 B's.
STEP 3: Let's count the total number of game arrangements.
Since either team can win (or lose) the series the same number of ways, we can take advantage of the available symmetry and only look at possibilities that lead to team A winning the series and then mulitply it by 2.
Since we have the restriction that A must win the last game, the rest of the series games will have exactly three A's from which to create our groups of games. The number of B's to choose from will equal the series length minus four.
Instead of calculating the probability for each series length, we can simply calculate the probability of the series going seven games, and then subtracting that from 1.
p(A winning in 7)
------A
How many words of six ordered letters can we have with three A's and three B's in it? The answer is 6!/3!/3! = 20
Each distinct order has a probability of 1/2^7 = 1/128 of occuring, so our total probability of A winning in seven games is 20/128 = 5/32, which means there is a 10/32 probability of either A or B winning in seven games.
p(less than 7) = 1 - p(seven) = 22/32 = 11/16
D
