Question:
Let \( X \) be a random variable having distribution function
\[
F(x) =
\begin{cases}
0, & x<1 \\
a, & 1 \leq x<2 \\
\frac{c}{2}, & 2 \leq x<3 \\
1, & x \geq 3
\end{cases}
\]
where \( a \) and \( c \) are appropriate constants. Let \( A_n = \left[ 1 + \frac{1}{n}, 3 - \frac{1}{n} \right] \), \( n \geq 1 \), and \( A = \bigcup_{i=1}^{\infty} A_i \). If \( P(X \leq 1) = \frac{1}{2} \) and \( E(X) = \frac{5}{3} \), then \( P(X \in A) \) equals _________ (round off to 2 decimal places).
Let \( X \) be a random variable having distribution function
\[ F(x) = \begin{cases} 0, & x<1 \\ a, & 1 \leq x<2 \\ \frac{c}{2}, & 2 \leq x<3 \\ 1, & x \geq 3 \end{cases} \]
where \( a \) and \( c \) are appropriate constants. Let \( A_n = \left[ 1 + \frac{1}{n}, 3 - \frac{1}{n} \right] \), \( n \geq 1 \), and \( A = \bigcup_{i=1}^{\infty} A_i \). If \( P(X \leq 1) = \frac{1}{2} \) and \( E(X) = \frac{5}{3} \), then \( P(X \in A) \) equals _________ (round off to 2 decimal places).Show Hint
To solve for probabilities in such distribution problems, use the properties of the cumulative distribution function and apply the given information systematically.
Updated On: Dec 29, 2025
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Correct Answer: 0.32
Solution and Explanation
We are given the distribution function and the probabilities \( P(X \leq 1) = \frac{1}{2} \) and \( E(X) = \frac{5}{3} \). From the distribution function, we can solve for the values of \( a \) and \( c \). Using the properties of the cumulative distribution function and the expected value formula, we can determine that:
\[
P(X \in A) = 0.32.
\]
Thus, the value is \( 0.32 \).
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