Probability & Statistics Ch.2 Section 3 Question 6

· Mohammad-Ali Bandzar

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 Z∼Geometric(θ)Z \sim \text{Geometric}(\theta). Compute P(5≤Z≤9)P(5 \leq Z \leq 9).

Solution

From Example 2.3.4 in the textbook, we know:

pZ(z)=(1−θ)zθp_Z(z) = (1-\theta)^z \theta

Now we add up the geometric probabilities for Z = 5, 6, 7, 8, 9:

P(5≤Z≤9)=(1−θ)5θ+(1−θ)6θ+(1−θ)7θ+(1−θ)8θ+(1−θ)9θP(5 \leq Z \leq 9) = (1-\theta)^5\theta + (1-\theta)^6\theta + (1-\theta)^7\theta + (1-\theta)^8\theta + (1-\theta)^9\theta

This can be factored as:

P(5≤Z≤9)=θ(1−θ)5[1+(1−θ)+(1−θ)2+(1−θ)3+(1−θ)4]P(5 \leq Z \leq 9) = \theta(1-\theta)^5 \left[1 + (1-\theta) + (1-\theta)^2 + (1-\theta)^3 + (1-\theta)^4\right]

That is the probability that Z will be between 5 and 9 inclusively. Note that Z can only be a non-negative integer as it represents the number of failures before the first success.