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Chapter 1: Problem 2

Let the space of the random variable \(X\) be \(\mathcal{C}=\\{x: 0

Short Answer

Expert verified
\(P_{X}\left(C_{2}\right) = \frac{5}{8}\). So, certainly, \(P_{X}\left(C_{2}\right) \leq \frac{5}{8}\). Thus, the condition is always true.

Step by step solution

01

Understanding the Space of the Random Variable

The space of the random variable \(X\) is \(\mathcal{C}=\{x: 0<x<10\}\). So the random variable \(X\) can take any value between 0 and 10. The set \(C_{1}=\{x: 1<x<5\}\) represents all values such that 1 < x < 5, and \(P_{X}\left(C_{1}\right)=\frac{3}{8}\), which means the probability of the set \(C_{1}\) is \(\frac{3}{8}\).
02

Deriving \(P_{X}\left(C_{2}\right)\)

The set \(C_{2}\) is defined as \(C_{2}=\{x: 5 \leq x < 10\}\) and it represents all values such that 5 <= x < 10. Since \(C_{1}\) and \(C_{2}\) are disjoint sets (they don't have any common elements) and they together form the whole space of the random variable \(X\), we know that \(P_{X}(\mathcal{C}) = P_{X}(C_{1} \cup C_{2})\). Since \(C_{1}\) and \(C_{2}\) are disjoint, the probability of their union is the sum of their probabilities, so \(P_{X}(\mathcal{C}) = P_{X}(C_{1}) + P_{X}(C_{2})\).
03

Solving for \(P_{X}\left(C_{2}\right)\)

Knowing that \(P_{X}(\mathcal{C}) = 1\), we substitute the known values: \(1 = \frac{3}{8} + P_{X}(C_{2})\). Solving for \(P_{X}(C_{2})\), we get \(P_{X}(C_{2}) = 1 - \frac{3}{8} = \frac{5}{8}\).

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Most popular questions from this chapter

Let \(f(x)=1 / x^{2}, 1

Let \(X\) be a random variable such that \(E\left[(X-b)^{2}\right]\) exists for all real \(b\). Show that \(E\left[(X-b)^{2}\right]\) is a minimum when \(b=E(X)\).

If \(C_{1}\) and \(C_{2}\) are independent events, show that the following pairs of events are also independent: (a) \(C_{1}\) and \(C_{2}^{c}\), (b) \(C_{1}^{c}\) and \(C_{2}\), and (c) \(C_{1}^{c}\) and \(C_{2}^{c}\). Hint: In (a), write \(P\left(C_{1} \cap C_{2}^{c}\right)=P\left(C_{1}\right) P\left(C_{2}^{c} \mid C_{1}\right)=P\left(C_{1}\right)\left[1-P\left(C_{2} \mid C_{1}\right)\right]\). From independence of \(C_{1}\) and \(C_{2}, P\left(C_{2} \mid C_{1}\right)=P\left(C_{2}\right)\).

Let \(0

Let \(C_{1}\) and \(C_{2}\) be independent events with \(P\left(C_{1}\right)=0.6\) and \(P\left(C_{2}\right)=0.3\). Compute (a) \(P\left(C_{1} \cap C_{2}\right) ;(b) P\left(C_{1} \cup C_{2}\right) ;\) (c) \(P\left(C_{1} \cup C_{2}^{c}\right)\).
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