<p>Joe is going to work between 3 - 5 hours this Saturday. Bob is going to work between 2 - 7 hours this Saturday. What is the probability that Bob will work at least as many hours as Joe.</p>
<p>WRONG:
Joe can work 1 or 2 hours
Bob can work 1, 2, 3, 4, 5 hours</p>
<p>The only way Joe works more hours than Bob if he works 2 hours and Bob works 1 hour. The odds of this are: 1/2 (odds Joe works 2 hours) x 1/5 (odds Bob works 1 hour) = 1/10</p>
<p>Therefore the probably Bob works atleast as much as Joe is 1 - 1/10 = 9/10</p>
<p>This is of course, assuming they both only work a whole number of hours and that they are equally likely to work any number of hours between 1-2, and 1-5.</p>
<hr>
<p>EDIT: Gee, I might want to learn read.</p>
<p>For some reason I thought your post said “between 3-5” as in 3 o’clock to 5 o’clock, not 3-5 hours. Sorry.</p>
<p>This is the correct probability:
Joe can work 3, 4, or 5 hours
Bob can work 2, 3, 4, 5, 6, or 7 hours</p>
<p>The only way Joe works more than Bob is if:
A. Joe works 3 hours and Bob works 2 hours
B. Joe works 4 hours and Bob works 2 OR 3 hours
C. Joe works 5 hours and Bob works 2, 3, OR 4 hours.</p>
<p>The odds of A happeneing are (1/3) (odds of Joe working 3 hours) x (1/7) (odds of Bob working 2 hours)</p>
<p>Odds of B are (1/3)x(2/7)</p>
<p>Odds of C are (1/3)x(3/7)</p>
<p>which is 1/21 + 2/21 + 1/7 = 2/7</p>
<p>The odds Bob works more than Joe are then 1 - 2/7 = 5/7.</p>
<p>oops, i meant to ask </p>
<p>Joe is going to work between 3 - 5 hours this Saturday. Bob is going to work between 2 and 7 hours this Saturday. What is the probability that Bob will work at least as many hours as Joe.</p>
<p>Not 2 - 7 hours, does that make a difference, it’s student produced, and the answer says it’s 60%</p>
<p>Well, one mistake I made is making Bob’s odds out of 7 - when 2, 3, 4, 5, 6 ,7 is only 6 possibilities. Therefore it should really be 1/3 * 1/6, 1/3 * 2/6, 1/3 * 3/6, which when subtracted from 1 would give you 2/3.</p>
<p>Are you sure the answer isn’t 2/3 (66.7%)?</p>
<p>it’s exactly 0.6</p>
<p>ObsessedAndre:
Does the #hours worked have to be integer?</p>
<p>nvm I figured it out thanks anyway</p>
<p>I would have done it the same way Andre did it and got 2/3, but since that’s not right I thought maybe they didn’t have to work a whole number of hours and that they worked “between,” so not inclusive of. </p>
<p>Joe works between 3-4 hours, 4 hours, or between 4-5 hours.
Bob works between 2-3,3, 3-4,4, 4-5,5, 5-6,6, 6-7 hours.
3 X 9 = 27</p>
<p>Joe more than Bob:
J - 3-4 B 2-3 or 3
J - 4 B 2-3, 3, 3-4
J - 4-5 B 2-3,3,3-4,4</p>
<p>Probabilites would be: 1/3 * 2/9 = 2/27 + 3/27 + 4/27 = 9/27 = 1/3
1- 1/3 = 2/3. Hmm. Even then I get 2/3.</p>
<p>Hmm… I guess they don’t necessarily have to be integers but then I don’t see how you’d figure it out.</p>
<p>r3n-_-, would you like to post the solution?</p>
<p>obsessedAndre, what I just posted is if its not integer number of hours. Yeah, can you post the solution r3n?</p>
<p>i would also like to see the solution. must me some simple reason why the answer is .6 instead of 2/3.</p>
<p>The problem with evilmonkey’s distribution is that it’s still quantized. Just because Joe worked “between 3-4 hours” and Bob worked “between 3-4 hours” does not mean they worked the same amount of time: Joe could’ve worked 3 hours, 20 minutes; Bob could’ve worked 3 hours, 19 minutes.</p>
<p>Here’s how I did it:</p>
<p>Let x be the number of hours Joe works and let y be the number of hours Bob works. Graph this on a coordinate plane.</p>
<p>x could be from 3 to 5
y could be from 2 to 7</p>
<p>Ergo, the feasible region is a rectangle from (3, 2) to (5, 7) with area 10 (2 * 5).</p>
<p>For Bob to work more than Joe, y must be greater than x. Graph the line y = x on the chart, and find that it intersects the rectangle at (3,3) and (5,5). The area of the region fitting the constraints that (y > x) and that (2 < y < 7) and that (3 < x < 5) is therefore 6.*</p>
<p>6/10</p>
<p>I don’t ever recall learning this technique in school. I’ve used it multiple times on math team for continuous probability distributions, though.</p>
<p>*Area bounded by coordinate pairs (x,y) can be found by the Shoelace theorem (which is easily programmable) or you can look at the problem and do it by splitting the region into a triangle and a rectangle.</p>
<p>Thanks. That makes sense.</p>
<p>OctalC0de:
That’s neat! I don’t recall seeing this method being taught, either - it’s intuitive, and easy to grasp.</p>
<p>However, with >= 3 probabilistic events, you start having to compare 3-D volumes, which may not be that easy to visualize… :)</p>
<p>was this a sat question?</p>