Solution 2: Bertrand's Problem.

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(a)

1) Show D = cos (A)

Consider the triangle who has side D and whose hypotenuse is a part of the x-axis of length 1. Then cos (A) = D / 1 = D.

2) Note that the triangle whose height is Y and whose base is X has a hypotenuse that is the radius of the circle. So, X² + Y² = 1. Use the pythagorean theorem for the following triangle to solve for x.

In order to calculate the height of the pink triangle, use the distance formula for the points ( X, Y ) amd ( 1, 0 ) and divide by two. Using the pythagorean theorem, we have:

3) Use the equation X² + Y² = 1. Subsitute 1-2D² for X. Y = 2D (1-D²)^1/2.

4) Show A is also the angle between the chord and the tangent line to the circle at the point (0,1). Consider the isoceles triangle inscribed in the circle.

(b)

1) Consider the triangle in the figure above. Because D = 1/2 and the hypotenuse is 1, we can use the pythagorean theorem and show that the remaining side is . The triangle is then a 30-60-90 and A = /3.

2) Similarly since A = /3 and the hypotenuse is 1, it follows that D = 1/2.

3) Use the formula X = 2D² - 1.

4) Use the formula Y = 2D (1-D²)^1/2.

(c)

The circle has radius 1, so D is a point chosen at random from [0,1]. We know 0 < D < 1/2, so the probability we seek is (1/2) / 1 = 1/2.

(d)

/2 - /3 = /6, which is the event space. The probability is

Problem 2

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