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Chapter 1: Problem 6

Let the probability set function of the random variable \(X\) be \(P_{X}(D)=\) \(\int_{D} f(x) d x\), where \(f(x)=2 x / 9\), for \(x \in \mathcal{D}=\\{x: 0

Short Answer

Expert verified
The probabilities are \(P_{X}(D_{1}) = 1/9\), \(P_{X}(D_{2}) = 7/9\), and \(P_{X}(D_{1} \cup D_{2}) = 8/9.\)

Step by step solution

01

Compute \(P_{X}(D_{1})\)

First compute the integral of \(f(x)\) over the interval given by \(D_1\). This is given by \(\int_{D_1} f(x) d x\)= \(\int_{0}^{1} 2x/9 dx\). The antiderivative of \(2x/9\) is \(x^2/9\), so the result of the integral is \((1^2/9 - 0^2/9) = 1/9.\)
02

Compute \(P_{X}(D_{2})\)

Now compute the integral of \(f(x)\) over the interval given by \(D_2\). This is given by \(\int_{D_2} f(x) d x\)= \(\int_{2}^{3} 2x/9 dx\). The antiderivative of \(2x/9\) is \(x^2/9\), so the result of the integral is \((3^2/9 - 2^2/9) = 7/9.
03

Compute \(P_{X}(D_{1} \cup D_{2})\)

Since \(P_{X}(D_{1} \cup D_{2}) = P_{X}(D_{1}) + P_{X}(D_{2})\) in this case because \(D_1\) and \(D_2\) do not intersect, we can add the results from Step 1 and Step 2 to get \(1/9 + 7/9 = 8/9.

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Most popular questions from this chapter

After a hard-fought football game, it was reported that, of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt both an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.

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