# 2006 iTest Problems/Problem 16

## Problem

The Minnesota Twins face the New York Mets in the 2006 World Series. Assuming the two teams are evenly matched (each has a $.5$ probability of winning any game) what is the probability that the World Series (a best of 7 series of games which lasts until one team wins four games) will require the full seven games to determine a winner?

$\text{(A) }\frac{1}{16}\qquad \text{(B) }\frac{1}{8}\qquad \text{(C) }\frac{3}{16}\qquad \text{(D) }\frac{1}{4}\qquad \text{(E) }\frac{5}{16}\qquad$

$\text{(F) }\frac{3}{8}\qquad \text{(G) }\frac{5}{32}\qquad \text{(H) }\frac{7}{32}\qquad \text{(I) }\frac{9}{32}\qquad \text{(J) }\frac{3}{64}\qquad \text{(K) }\frac{5}{64}\qquad$

$\text{(L) }\frac{7}{64}\qquad \text{(M) }\frac{1}{2}\qquad \text{(N) }\frac{13}{32}\qquad \text{(O) }\frac{11}{32}\qquad \text{(P) }\text{none of the above}$

## Solution

In order for the World Series to have seven games, each team must win exactly three times. The last game does not matter. There are $\tbinom{6}{3} = 20$ possibilities, and since all of the possibilities involve each team winning three times, the probability of having 7 games in the World Series is $20 \cdot \left( \frac12 \right)^6 = \boxed{\textbf{(E) }\frac{5}{16}}$.

## See Also

 2006 iTest (Problems, Answer Key) Preceded by:Problem 15 Followed by:Problem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10