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Chapter 1: Problem 14

Let \(X\) have the pdf \(f(x)=2 x, 0

Short Answer

Expert verified
The probability that \(X\) is at least \(\frac{3}{4}\) given that \(X\) is at least \(\frac{1}{2}\) is \(\frac{7}{12}\).

Step by step solution

01

Identify the probability formula for a continuous random variable

The continuous random variable \(X\) has the pdf \(f(x)=2x\) for \(0 < x < 1\), and zero elsewhere. The conditional probability \(P(A|B)\) can be found using the ratio of probabilities \(P(A \cap B) / P(B)\). In this exercise, event \(A\) corresponds to \(X \geq \frac{3}{4}\), and event \(B\) corresponds to \(X \geq \frac{1}{2}\).
02

Calculate \(P(A)\) and \(P(B)\)

The probabilities \(P(A)\) and \(P(B)\) are obtained by integrating the pdf \(f(x)\) over the desired intervals. \[P(A) = \int_{3/4}^{1} f(x) dx = \int_{3/4}^{1} 2x dx = [x^2]_{3/4}^{1} = 1 - (3/4)^2 = 1 - 9/16 = 7/16\] \[P(B) = \int_{1/2}^{1} f(x) dx = \int_{1/2}^{1} 2x dx = [x^2]_{1/2}^{1} = 1 - (1/2)^2 = 1 - 1/4 = 3/4\]
03

Compute the conditional probability

Now, the conditional probability \(P(A|B)\) can be computed using the formula \(P(A \cap B) / P(B)\). In this case, \(P(A \cap B) = P(A)\) since \(A\) is included in \(B\). Hence, the conditional probability is \[P(A|B) = P(A) / P(B) = (7/16) / (3/4) = (7/16) * (4/3) = 7/12\]. This is the probability that \(X\) is at least \(\frac{3}{4}\) given that \(X\) is at least \(\frac{1}{2}\).

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