The future lifetimes (in months) of two components of a machine have the following joint density function:

Determine which of the following represents the probability that both components are still functioning 20 months from now.

• [math]\frac{6}{125000}\int_{0}^{20}\int_{0}^{20}(50-x-y)\,ds\,dt[/math]
• [math]\frac{6}{125000}\int_{20}^{30}\int_{20}^{50-x}(50-x-y)\,ds\,dt[/math]
• [math]\frac{6}{125000}\int_{20}^{30}\int_{20}^{50-x-y}(50-x-y)\,ds\,dt[/math]
• [math]\frac{6}{125000}\int_{20}^{50}\int_{20}^{50-x}(50-x-y)\,ds\,dt [/math]
• [math]\frac{6}{125000}\int_{20}^{50}\int_{20}^{50-x-y}(50-x-y)\,ds\,dt [/math]

An insurance company insures a large number of drivers. Let [math]X[/math] be the random variable representing the company’s losses under collision insurance, and let [math]Y[/math] represent the company’s losses under liability insurance. [math]X[/math] and [math]Y[/math] have joint density function

Calculate the probability that the total company loss is at least 1.

• 0.33
• 0.38
• 0.41
• 0.71
• 0.75

Let [math]T_1[/math] and [math]T_2[/math] represent the lifetimes in hours of two linked components in an electronic device. The joint density function for [math]T_1[/math] and [math]T_2[/math] is uniform over the region defined by

where [math]L[/math] is a positive constant. Determine the expected value of the sum of the squares of [math]T_1[/math] and [math]T_2[/math].

• [math]\frac{L^2}{3}[/math]
• [math]\frac{L^2}{2}[/math]
• [math]\frac{2L^2}{3}[/math]
• [math]\frac{3L^2}{4}[/math]
• [math]L^2[/math]

Let [math]X[/math] and [math]Y[/math] be continuous random variables with joint density function

Let [math]g[/math] be the marginal density function of [math]Y[/math]. Determine which of the following represents [math]g[/math].

• [[math]]g(y) = \begin{cases} 15y, \, 0 \lt y \lt 1 \\ 0, \, \textrm{Otherwise.} \end{cases}[[/math]]
• [[math]]g(y) = \begin{cases} \frac{15y^2}{2}, \, x^2 \lt y \lt x \\ 0, \, \textrm{Otherwise.} \end{cases}[[/math]]
• [[math]]g(y) = \begin{cases} \frac{15y^2}{2}, \, 0 \lt y \lt 1 \\ 0, \, \textrm{Otherwise.} \end{cases}[[/math]]
• [[math]]g(y) = \begin{cases} 15y^{3/2}(1-y^{1/2}), \, x^2 \lt y \lt x \\ 0, \, \textrm{Otherwise.} \end{cases}[[/math]]
• [[math]]g(y) = \begin{cases} 15y^{3/2}(1-y^{1/2}), \, 0 \lt y \lt 1\\ 0, \, \textrm{Otherwise.} \end{cases}[[/math]]

A device contains two circuits. The second circuit is a backup for the first, so the second is used only when the first has failed. The device fails when and only when the second circuit fails. Let [math]X[/math] and [math]Y[/math] be the times at which the first and second circuits fail, respectively. [math]X[/math] and [math]Y[/math] have joint probability density function

Calculate the expected time at which the device fails.

• 0.33
• 0.50
• 0.67
• 0.83
• 1.50

Two insurers provide bids on an insurance policy to a large company. The bids must be between 2000 and 2200. The company decides to accept the lower bid if the two bids differ by 20 or more. Otherwise, the company will consider the two bids further. Assume that the two bids are independent and are both uniformly distributed on the interval from 2000 to 2200.

Calculate the probability that the company considers the two bids further.

• 0.10
• 0.19
• 0.20
• 0.41
• 0.60

The proportion [math]X[/math] of yearly dental claims that exceed 200 is a random variable with probability density function

Calculate [math]\operatorname{Var}[X/(1 – X)][/math].

• 149/900
• 10/7
• 6
• 8
• 10

Skateboarders A and B practice one difficult stunt until becoming injured while attempting the stunt. On each attempt, the probability of becoming injured is [math]p[/math], independent of the outcomes of all previous attempts.

Let [math]F(x, y)[/math] represent the probability that skateboarders A and B make no more than [math]x[/math] and [math]y[/math] attempts, respectively, where [math]x[/math] and [math]y[/math] are positive integers. It is given that [math]F(2, 2) = 0.0441[/math].

Calculate [math]F(1, 5)[/math].

• 0.0093
• 0.0216
• 0.0495
• 0.0551
• 0.1112

Random variables [math]X[/math] and [math]Y[/math] have joint distribution

Let [math]a[/math] be the value that minimizes the variance of [math]X.[/math]

Calculate the variance of [math]Y[/math].

• 2/5
• 8/15
• 16/25
• 2/3
• 7/10

An insurance company sells two types of auto insurance policies: Basic and Deluxe. The time until the next Basic Policy claim is an exponential random variable with mean two days. The time until the next Deluxe Policy claim is an independent exponential random variable with mean three days. Calculate the probability that the next claim will be a Deluxe Policy claim.

• 0.172
• 0.223
• 0.400
• 0.487
• 0.500