Alexa was coming up to her final exams, which were set across 18 days. Each of the 18 days had 2 sessions, a morning and an afternoon session. Alexa had to complete 6 exams altogether.
When the exam timetable was released, Alexa was very disappointed that she had 2 days on which she had to complete 2 exams, one in the morning and one in the afternoon, on that same day.
Determine the probability that Alexa would find her 6 exams set on just 4 days.
My method was a bit more complicated than Charlie's but they agree:
Imagine choosing 6 objects out of a set of 36 which are paired, one by one. Each choosing is either a Hit or a Miss. Hit if it makes a pair, Miss if it doesn't.
The first choice must be a Miss. There cannot be more Hits than Misses at any point.
The value of a Hit equals the number of available Misses before it.
For there to be exams on 4 days, we must have two Hits and four Misses.
Here they are, with their values:
MHMHMM 1x1=1
MHMMHM 1x2=2
MHMMMH 1x3=3
MMHHMM 2x1=2
MMHMHM 2x2=4
MMHMMH 2x3=6
MMMHHM 3x2=3
MMMHMH 3x3=9
MMMMHH 4x3=12
For a grand total 45.
The probability is thus 45*(36*34*32*30)/(36*35*34*33*32*21)=
90/2387
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I may as well share the rest of the distribution solved my way:
All tests on 3 days:
MHMHMH 1x1x1=1
MHMMHH 1x2x1=2
MMHHMH 2x1x1=2
MMHMHH 2x2x1=4
MMMHHH 3x2x1=6
Grand total 15.
15*(36*34*32)/(36!/30!)=1/2387
Tests on 5 days:
MHMMMM 1
MMHMMM 2
MMMHMM 3
MMMMHM 4
MMMMMH 5
Grand total 15.
15*(36*34*32*30*28)/(36!/30!)=840/2387
Tests all on different days
MMMMMM 1
(36*34*32*30*28*26)/(36!/30!)=1456/2387
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Posted by Jer
on 2023-12-15 14:39:09 |
More complicated solution / full dist.
