The probability that s+d > 3 and sd > 3 is 0.03.
Solution
Plot the inequalities
The region -5 ≤ s,d ≤ 5 is the square shaded in grey.
The region s + d > 3 is the region Q shaded to the right of the straight line.
The region sd > 3 is the region R shaded to the right of the curve d = 3/s.
Find the intersection of the three regions
From the figure, the region satisfying all the above three inequalities is the region to the left of the curve d = 3/s, bounded by the square region, i.e. the region R.
Probability of region R
The required probability is the Geometric probability of the intersection region R. It is calculated as
P(R) = ar(region R) / ar(square region P).
Calculate the areas of the regions
ar(region R) = area of the rectangle to the right in the first quadrant formed by dropping a vertical from point F - area under the curve d = 3/s in the first quadrant
[tex]\[\Rightarrow \;\; \mathrm{ar(\mathbf{R}) =} \int_{3/5}^5\frac{3}{s}ds = 25-3-3\ln\frac{25}{3} = 15.64.\][/tex]
ar(region P) = 25 × 25 = 625.
Calculate P(R)
The probability of the region R, P(R) = 15.64 / 625 = 0.025.
Rounding it to the hundredth place of decimal, P(R) = 0.03.
The probability that s+d >3 and sd>3, where -5 < s,d < 5, is 0.03.
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Question
What is the probability that s + d > 3 and sd > 3, where -5 ≤ s,d ≤5? Write your answer as a decimal rounded to the hundredth place.
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