Question:
Let X1 and X2 be two independent and identically distributed discrete random variables having the probability mass function
\(f(x) = \begin{cases} (\frac{1}{2})^x, & x=1,2,3,... \\ 0, & \text{otherwise. } \end{cases}\)
Then P(min{X1 , X2} ≥ 5) equals
Let X1 and X2 be two independent and identically distributed discrete random variables having the probability mass function
\(f(x) = \begin{cases} (\frac{1}{2})^x, & x=1,2,3,... \\ 0, & \text{otherwise. } \end{cases}\)
Then P(min{X1 , X2} ≥ 5) equals
\(f(x) = \begin{cases} (\frac{1}{2})^x, & x=1,2,3,... \\ 0, & \text{otherwise. } \end{cases}\)
Then P(min{X1 , X2} ≥ 5) equals
Updated On: Oct 1, 2024
- \(\frac{1}{256}\)
- \(\frac{1}{512}\)
- \(\frac{1}{64}\)
- \(\frac{9}{256}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A) : \(\frac{1}{256}\).
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