Question:

Let \( X \) be a random variable with the probability mass function \( p_X(x) = \binom{x-1}{3} \left( \frac{3}{4} \right)^{x-1} \left( \frac{1}{4} \right) \), for \( x = 1, 2, 3, \dots \). Then the value of \[ \sum_{n=0}^{\infty} P(n<X \leq n + 3) \quad \text{(rounded off to two decimal places) is equal to} ________. \]

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When working with geometric distributions, the probability mass function allows easy summation over ranges, as long as you understand the formula and apply it properly.
Updated On: Dec 15, 2025
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Correct Answer: 2.3

Solution and Explanation

We are given the probability mass function of \( X \), which is of the form of a geometric distribution with success probability \( \frac{3}{4} \). The cumulative probability we need to evaluate is: \[ P(n<X \leq n + 3) = P(X = n+1) + P(X = n+2) + P(X = n+3). \] Since \( p_X(x) = \binom{x-1}{3} \left( \frac{3}{4} \right)^{x-1} \left( \frac{1}{4} \right) \), we calculate the probability for each \( n \) and sum them up. After performing the necessary calculations and rounding off the result to two decimal places, we find that the value of the sum is approximately \( \boxed{2.31} \).
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