The Mathematics department has got 6 tickets for Mark O\'Connor at the Harriet S
ID: 3301846 • Letter: T
Question
The Mathematics department has got 6 tickets for Mark O'Connor at the Harriet S. Jorgensen Theater The seats have been picked randomly from a wing with 5 rows and 10 seats each. What is the probability that (a) all seats are in the same row? (b) exactly four of them are in the same row? (The remaining two are supposed to be single spots each in a different row.) What is the probability that among the 13 members of the UConn Husky women's basketball team, exactly 5 of them have birthday in the same month?Explanation / Answer
1. (a) In each row, there are 10 seats. We can select 6 seats in each row in
10C6 ways
Since there are 5 rows, we can do this in
5*10C6
= 5 * 210
= 1050 ways
Total number of ways of selecting 6 seats out of 50 seats
= 50C6
= 15890700
Probability that all seats are in the same row
= 1050 / 15890700
= 0.000066
(b) The number of ways of selecting 4 seats in a row = 5 * 10C4 = 5 * 210 = 1050
The remaining two seats (two rows) can be selected in 4C2 * 10 = 60 ways
=> Total number of ways to assign seats with four in the same row = 1050 * 60 = 63000
Probability = 63000 / 15890700
= 0.003965
2. The five members can be chosen in 13C5 = 1287 ways
The first member can have birthday in any month
Probability that the second member has birthday in the same month = 1/12
Probability that the third memeber has birthday in the same month = 1/12
Probability that the fourth member has birthday in the same month = 1/12
Probability that the fifth member has birthday in the same month = 1/12
=> Probability that all five members have birthday in the same month = (1/12)4
Probability that the remaining eight members do not have birthday in the same month = (11/12)8
=> Probability that exactly five members have birthday in the same month = 13C5 * (1/12)4 (11/12)8
= 1287 * 0.000024
= 0.0309
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