Probability of a-b > 3 in Randomly Selected Intervals

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The problem involves finding the probability that the difference a - b is greater than 3, with a randomly selected from the interval [0, 4] and b from [-3, 0]. The maximum value of a - b can reach 7, as calculated from the intervals. To determine the probability, one must visualize the possible values of a and b in a two-dimensional plane, forming a rectangle. The next step is to identify the area of the rectangle where the condition a - b ≥ 3 holds true and calculate the fraction of this area relative to the total rectangle. This approach will yield the desired probability.
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Homework Statement



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.

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The Attempt at a Solution



So, 0<=a<=4 and -3<=b<=0 then 0<=a-b<=7. So is the probability that a-b is greater than 3, 4/8? There are 8 chances and only 4 probabilities that it could happen, 4,5,6,7.
 
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Your choices aren't limited to integers. Draw the region of possible values of a and b in a two dimensional plane where one axis is a and the other is b. It's a rectangle, right? What fraction of that rectangle will satisfy a-b>=3? Find the line that bounds that region.
 
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