What is the probability?
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M$4 Answers
5/7 * 4/6 * 3/5 = (5 * 4 * 3)/(7 * 6* 5) = 2/7 or about 28.57%.
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M$The 1st draw is a non-weekend is 5/7 -- as there 5 non-weekend days
The 2nd draw is a non-weekend is 4/7
The 3rd draw is a non-weekend is 3/7
So the required probability is (5/7)*(4/7)*(3/7) =
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M$
There is a 36% chance of this happening (rounding for clarity). Or looking at it the other way, there is a 64% chance your weekend will be spoiled. Don't leave scheduling to chance!
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M$ordered or unordered; the answers will be the same. For ordered samples, the
size of the sample space is 7P3 = 7 • 6 • 5 = 210. If A is the event ‘no lectures at
weekends’, then A occurs precisely when all three days drawn are weekdays; so
|A| = 5P3 = 5 • 4 • 3 = 60. Thus, P(A) = 60/210 = 2/7.
If we decided to use unordered samples instead, the answer would be 5C3/7C3,
which is once again 2/7.
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M$
