Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
This is a true story! Alice 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. Alice had to complete 6 exams altogether.
When the exam timetable was released, Alice cried, "I'm screwed! I've got two doublers! What are the chances of that?" By this Alice meant that she was 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. What is the probability that Alice would find her 6 exams set on just 4 days?
HintTry counting the number of ordered selections.
STEP 1: Count the number of ways of arranging the 6 exams into 4 days. The first day can be selected 6P2 (=30) ways. The second day can be selected 4P2 (=12) ways. The third day can be selected 2x2 (=4) ways and the last day can be selected 1x2 (=2 ) ways. 30x12x4x2 = 2880 ways.
STEP 2: Count the number of arrangements of the 4 days (2 double days and 2 singles) = 4!/2!/2! = 6 ways.
STEP 3: 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.
STEP 4: The number of ways of allocating the exams in the given pattern is 2880 x 6 x 3060 = 52,876,800.
STEP 5: Calculate the sample space. The number of ways that 6 exams can be arranged in the 36 available sessions is 36P6 = 1,402,410,240.
STEP 6: The probability is thus: 52,876,800/1,402 410,240 = 90/2387 or 0.0377.
Alice has just under 4% chance of getting this pattern, so maybe she has a right to feel unlucky!
May 25, 2006
|This has come up before. I think there is an easier answer. I'm just glad it's not my probem any more.|
May 25, 2006
|Nice job, Jimbo! Just hope you haven't jinxed me with this teaser... |
Oct 18, 2006
|But true randomness is not in effect. Factors like class size and grade level may skew the results.|
Oct 21, 2006
|I calculated about a 1% chance- is my logic wrong-|
36 total sessions, so the chance of the first four exams being on separate days = 1 * 34/35 * 32/34 * 30/33, then the chance of the 5th exam being on one of those 4 days and the 6th exam being on one of the 3 remaining days is 4/32 * 3/31.
So 1 * 34/35 * 32/34 * 30/33 * 4/32 * 3/31 = about 1.005%.
Oct 22, 2006
|I think you will find that you have omitted to take into account that there are several different orders that this could occur. For example the first exam could be Geography and the second one could be History. Using he same time slots but reversing the order woud be another way that the exams could be scgeduled but you have not counted these.|
Apr 25, 2007
|I dont get it. How are you to figure this out? Are all 6 of Alices exams offered in all 36 sessions? Are there any criteria for scheduling that would encourage bunching of exam dates? Are the exam sessions broken up by alphabetical order or subject matter? Science math first week, english, lit second week, pre-med third week.|
Are the days consecutive or do they have to be? And finally do you mean to say 4 days exactly or 4 or less days? she could have had 3 doubles which then throws the probabilities all off. What are the chances that her exams where scheduled in exactly the order they where as opposed to being scheduled in the same sessions but in different order?
Apr 28, 2007
|Each exam will be scheduled once only in the 36 available time slots. The question says find the probability of 4 days (exactly) not 4 or less. In 4 days there must be 2 doubles and 2 single exams to make the six exams. It is a tricky problem but a real one which makes students believe that they have been very unlucky or harshly treated. The question seeks to explore the probability to find if this is so or is it reasonably likely.|
Aug 14, 2008
|Made a computer simulation and got it to 2.9% so I believe your answer is wrong.|
Aug 14, 2008
|How many times did you run the simulation?|
Dec 08, 2008
|The problem can be solved more easily. First, since you don't care which test you take on each day, permutations can be eliminated from the problem space.|
This means that there are 36C6 = 1,947,792 ways to distribute the tests across 36 testing slots.
There are 18C2 = 153 ways to select which days you have two tests and 16C2 = 120 ways to select the days for the remaining two tests. Each day with one test can have the test in the morning or afternoon, so there are 2x2 = 4 ways to distribute the tests. This gives 153x120x4 = 73,440 ways to have two tests on exactly one day. (This excludes the 18C3 = 816 ways to get three "doubles".)
The probability is then 73440/1947792 = 90/2387 = 0.0377...
Dec 08, 2008
|Thanks Guru! That's a very elegant solution. I wish I had come up with that! |
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