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(a)
Let x and y denote the number of minutes after 12:00 PM that Mr. Spraggins' and Mrs. Taylor's arrivals occur, respectively. Thus, if Mr. Spraggins arrives at x minutes after 12:00 PM, then the two will see each other if and only if Mrs. Taylor arrives less than fifteen minutes before or after Mr. Spraggins; that is, iff
x - 15 < y < x + 15.
Below is a diagram of the event space, similar to Problem 0.
Thus, the probability of a successful outcome is (8100-2*2812.5)/8100, which is approximately 0.31.
(b)
Let x and y be as they were above. Once again, our protagonists will see each other if and only if
The diagram below represents the event space, that is, all instances when the above equation is satisfied. Note that this is a discrete case and thus has a finite number of points. We are unable to compare areas, so we instead determine the number of points/events in the sample space compared to the number of points/events in the event space. Thus, the probability of a successful outcome is the ratio of the number of points in the strip to the total number of points in the sample space. Observe that only those points which are multiples of five exist, thus the sample space is a 19 by 19 grid.
Thus, the entire grid consists of 19*19 = 361 points. So the probability is (361 - 2*120) / 361, which is approximately 0.335.
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EXERCISES: The following exercises are from NCTM
- Problem 3 from the 6.1 Exercises on page 11.
- Problem 9 from the Exercises on page 26.
- Problem 14, part (a), from the Supplementary Exercises on page 31.
Solutions
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