UNO High School Problem of the Week Competition
Problem 20
Final exams took place over a span of 18 days. Each of the 18 days had a morning and an afternoon session. A girl had to take 6 exams total. She looked at the exam schedule and found she had two days in which she had to take two finals each, one in the morning and one in the afternoon. She thought to herself, "What are the chances of that happening?" What is the probability that she would have 6 exams set on just 4 days?
Show your work.
Solution
Count the number of ways of arranging the 6 exams into 4 days. The first day can be selected 6 tests in 2 time slots (=30) ways. The second day can be selected 4 tests in 2 time slots (=12) ways. The third day can be selected 2x2 (=4) ways and the last day can be selected 1x2 (=2) ways. 30x12x4x2 = 2880 ways.
Count the number of arrangements of the 4 days (2 double days and 2 singles) = 4!/2!/2! = 6 ways
Count the number of ways that the 4 days can be allocated to the 18 available days. Note that since every arrangement within the 4 days has already been counted, order will not be considered. 18C4 = 3060
The number of ways of allocating the exams in the given pattern is 2880 x 6 x 3060 = 52 876 800.
Calculate the sample space. The number of ways that 6 exams can be arranged in the 36 available sessions is 36 sessions in 6 days = 1,402,410,240.
The probability is thus, 52876 800/1 402 410 240 = 90/2387 or 0.0377.
The girls has just under 4% chance of getting this pattern.