$P(X = k) = (1-p)^{k-1} p, \quad k = 1, 2, 3, \ldots$ The Geometric Distribution $X \sim \text{Geo}(p)$ models the number of independent Bernoulli trials needed to obtain the first success, where each trial succeeds with probability $p$. Its mean is $E[X] = 1/p$ and variance is $\text{Var}(X) = (1-p)/p^2$. A key property is **memorylessness**: given that no success has occurred in the first $k$ trials, the remaining waiting time has the same geometric distribution. > [!example]- Waiting for the First Fraudulent Claim {💡 Example} > An adjuster reviews claims one at a time. Each claim is fraudulent with probability $p = 0.1$, independently. What is the probability the first fraudulent claim is the 5th one reviewed? > > > [!answer]- Answer > > $P(X = 5) = (1-0.1)^{5-1}(0.1) = (0.9)^4 \times 0.1 = 0.6561 \times 0.1 \approx 0.0656$