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The probability that the sum of two numbers x and y randomly chosen in the interval (0, 1) greater than 1 while the sum of the squares less than 1 is equal to
- a)2/π
- b)π/4
- c)π/6 - 1/2
- d)π/4 - 1/2
Correct answer is option 'D'. Can you explain this answer?
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To find the probability, we need to find the area of the region in the (x, y) plane where the sum of x and y is greater than 1 and the sum of the squares is less than 1.
Let's consider the first condition: x + y > 1. This represents the region above the line y = 1 - x in the (x, y) plane.
Now let's consider the second condition: x^2 + y^2 < 1.="" this="" represents="" the="" region="" inside="" the="" unit="" circle="" centered="" at="" the="" />
To find the area of the region that satisfies both conditions, we need to find the intersection of these two regions.
The line y = 1 - x intersects the unit circle at two points: (-0.707, 0.707) and (0.707, -0.707).
The area of the region that satisfies both conditions is the area of the unit circle minus the area of the triangle formed by the line and the x and y axes.
The area of the unit circle is πr^2 = π(1^2) = π.
The area of the triangle is (1/2)(1)(1) = 1/2.
Therefore, the area of the region that satisfies both conditions is π - 1/2.
Since the two numbers x and y are chosen randomly in the interval (0, 1), the total area of the possible outcomes is 1.
The probability that the sum of x and y is greater than 1 and the sum of the squares is less than 1 is therefore (π - 1/2)/1 = π - 1/2.
However, this expression cannot be simplified further and does not equal 2/.
Let's consider the first condition: x + y > 1. This represents the region above the line y = 1 - x in the (x, y) plane.
Now let's consider the second condition: x^2 + y^2 < 1.="" this="" represents="" the="" region="" inside="" the="" unit="" circle="" centered="" at="" the="" />
To find the area of the region that satisfies both conditions, we need to find the intersection of these two regions.
The line y = 1 - x intersects the unit circle at two points: (-0.707, 0.707) and (0.707, -0.707).
The area of the region that satisfies both conditions is the area of the unit circle minus the area of the triangle formed by the line and the x and y axes.
The area of the unit circle is πr^2 = π(1^2) = π.
The area of the triangle is (1/2)(1)(1) = 1/2.
Therefore, the area of the region that satisfies both conditions is π - 1/2.
Since the two numbers x and y are chosen randomly in the interval (0, 1), the total area of the possible outcomes is 1.
The probability that the sum of x and y is greater than 1 and the sum of the squares is less than 1 is therefore (π - 1/2)/1 = π - 1/2.
However, this expression cannot be simplified further and does not equal 2/.
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The probability that the sum of two numbers x and y randomly chosen in the interval (0, 1) greater than 1 while the sum of the squares less than 1 is equal toa)2/πb)π/4c)π/6 - 1/2d)π/4 - 1/2Correct answer is option 'D'. Can you explain this answer?
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