Probability, geometric series
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Probability, geometric series

[From: ] [author: ] [Date: 11-06-08] [Hit: ]
Why not, since 0=3, and the question ask when a-b is greater than 3. So, 4,5,......
On the real line (-inf, inf), point a resides in the interval 0<=a<=4, and point b is in the interval -3<=b<=0; otherwise, points a and b are selected at random within their intervals. Find the probability that a-b is greater than 3.

Answer:0<=a-b<=7, since a-b>=3 is 3/7th of the sample space, the required probability is 3/7.
Is this correct?
Why not, since 0<=a-b<=7 and a-b>=3, and the question ask when a-b is greater than 3. So, 4,5,6,7. That is 4/8.

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The answer should be 4/7 (by the way, I know I told you 3/7 last time, but it was like 2am when I posted it). This is because, in order for 0 ≤ a - b ≤ 7 and a - b ≥ 3 to both be satisfied, 3 ≤ a - b ≤ 7. This interval, [3, 7], has a length of 4, while the sample space, [0, 7], has a length of 7. Thus, the required probability is 4/7.

I hope this helps!
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keywords: series,geometric,Probability,Probability, geometric series
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