## Testing TimeProbability 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? ## AnswerSTEP 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! Hide ## What Next?
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