Probability & Statistics Ch.2 Section 3 Question 10

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let $X \sim \text{Geometric}(1/5)$ . Compute $P(X^2 \leq 15)$ .

Solution

From Example 2.3.4 in the textbook, we know:

p_X(x) = P(X=x) = (1-\theta)^x\theta

Where x represents the number of failures before a successful trial, and θ is the probability of success.

We plug in $\theta = \frac{1}{5}$ :

P(X=x) = \left(\frac{4}{5}\right)^x \cdot \frac{1}{5}

We want $P(X^2 \leq 15)$ :

X^2 \leq 15

X \leq \pm\sqrt{15}

Since X must be greater than or equal to zero, we ignore the negative root:

X \leq \sqrt{15} \approx 3.873

We round down to the nearest whole number (you can’t fail 0.87 trials):

X \leq 3

We add up the probability for k = 0, 1, 2, 3 failures:

P(X \leq 3) = \left(\frac{4}{5}\right)^0 \cdot \frac{1}{5} + \left(\frac{4}{5}\right)^1 \cdot \frac{1}{5} + \left(\frac{4}{5}\right)^2 \cdot \frac{1}{5} + \left(\frac{4}{5}\right)^3 \cdot \frac{1}{5}

= \frac{1}{5} + \frac{4}{25} + \frac{16}{125} + \frac{64}{625}

= \frac{125}{625} + \frac{100}{625} + \frac{80}{625} + \frac{64}{625}

P(X^2 \leq 15) = \frac{369}{625}