Question

Let the random variable \( X \) have uniform distribution on the interval \( \left( \frac{\pi}{6}, \frac{\pi}{2} \right) \). Then
\[ P(\cos X>\sin X) \] is

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{2} \)

Hint:

For uniform distributions, probabilities are determined by the length of the relevant intervals.

Answer 1

The Correct Option is B

Step 1: Define the uniform distribution.
The random variable \( X \) has a uniform distribution on the interval \( \left( \frac{\pi}{6}, \frac{\pi}{2} \right) \), so the probability density function (pdf) is: \[ f(x) = \frac{1}{\frac{\pi}{2} - \frac{\pi}{6}} = \frac{1}{\frac{\pi}{3}} = \frac{3}{\pi}, \quad \frac{\pi}{6} \leq x \leq \frac{\pi}{2} \] We are asked to find \( P(\cos X > \sin X) \).

Step 2: Solve the inequality \( \cos X > \sin X \).
We need to solve for \( X \) such that: \[ \cos X > \sin X \] Dividing both sides by \( \cos X \) (which is positive on the interval \( \left( \frac{\pi}{6}, \frac{\pi}{2} \right) \)): \[ 1 > \tan X \quad \Rightarrow \quad X < \tan^{-1}(1) \] Since \( \tan^{-1}(1) = \frac{\pi}{4} \), the inequality holds when \( X < \frac{\pi}{4} \).

Step 3: Calculate the probability.
The probability is the length of the interval where \( \cos X > \sin X \), i.e., from \( \frac{\pi}{6} \) to \( \frac{\pi}{4} \), divided by the total length of the interval \( \left( \frac{\pi}{6}, \frac{\pi}{2} \right) \). \[ P(\cos X > \sin X) = \frac{\frac{\pi}{4} - \frac{\pi}{6}}{\frac{\pi}{2} - \frac{\pi}{6}} = \frac{\frac{\pi}{12}}{\frac{\pi}{3}} = \frac{1}{4} \]

Step 4: Final Answer.
The probability \( P(\cos X > \sin X) = \frac{1}{4} \).

Final Answer: \( \frac{1}{4} \)