1)The World Series is a best-of-seven contest. (Two baseball teams play up to 7 games against each other. As soon as 1 team has won 4 games, the contest is over.) If teams A and B play in the series, a sequence of winners for the games played might be AAAA or ABBABAB, but could not be AAAAB.
How many different sequences of winners for the games of a World Series between Team A and Team B are possible?
Click here for the answers.
Answer
70
Suppose that a World Series is completed leading to an extended series where extra exhibition games are played (between the same two teams) after the series has been won. These exhibition games are always to be won by the team that lost the official World Series and are to continue until that losing team also has a total of 4 wins. In this way, if the results of the World Series read ABAABA, then the results of the extended series read ABAABABB.
The sequence of results for the extended series is just an arrangement of 4 copies of the letter A and 4 copies of the letter B. There are 8C4 = 8x7x6x5/4x3x2x1 = 70 such arrangements. (We may choose 4 positions for the letter A from
8 possible positions.)
Each sequence of results for the extended series determines exactly one possible sequence of results for the World Series (by omitting the exhibition games).
2) Two CUNY math professors arrange to meet to prepare a test. Each will arrive at a random time between 1:00 P.M. and 2:00 P.M. and will wait up to 15 minutes for the other before leaving.
What is the probability that the meeting takes place?
Click here for the answer.
Answer
7⁄16
Associate each arrival time between 1:00pm and 2:00pm with a real number
r such that 0 < r < 1. The number
r represents the part of the hour that has elapsed before the arrival.
In this way the arrival times of the two professors determine a coordinate pair (x,y) that corresponds to a point in the unit square with 0 < x < 1,0 < y < 1.
The pair (x,y) corresponds to a situation where the professors fail to meet if either y > x +
1⁄
4 or x > y +
1⁄
4 (see right).
In the first of these cases the point (x,y) lies in the triangle that lies above the line y = x +
1⁄
4. The area of this triangle is
1⁄
2 x
3⁄
4 x
3⁄
4 =
9⁄
32. Similarly, in the second case, the point (x,y) lies in the triangle with area
9⁄
32 that lies below the line x = y +
1⁄
4. We deduce that the probability that the professors do not meet is
9⁄
32 +
9⁄
32 =
9⁄
16. Therefore the probability that they do meet is
7⁄
16.
