An insurer covers 10 independent properties. Each of the 10 properties independently has probability 0.10 of a claim this year.
Calculate the probability that at least 2 of the 10 have a claim.
Pick a distribution, set its parameters and the event. You get the probability, the formula with your numbers in it, and the mean and variance.
Probability
0.2639
The chance of 2 or more.
| Binomial | Successes in n independent trials, each with probability p. |
|---|---|
| Poisson | Events in a fixed time at a steady average rate. |
| Geometric | Failures before the first success, each trial succeeding with probability p. |
| Negative binomial | Failures before the r-th success. |
| Hypergeometric | Successes in a sample drawn without replacement. |
| Uniform | Every value between a and b equally likely. |
| Exponential | A waiting time or loss with mean θ; memoryless. |
| Gamma | A sum of α exponential waits, each with mean θ. |
| Normal | The bell curve with mean μ and standard deviation σ. |
For a count, at least 2 includes 2, and more than 2 starts at 3. Most wrong answers on these questions are off by one value.
For a continuous variable a single value has probability zero, so below and at most are the same.
An insurer covers 10 independent properties. Each of the 10 properties independently has probability 0.10 of a claim this year.
Calculate the probability that at least 2 of the 10 have a claim.
The number of claims in a week has a Poisson distribution with mean 1.5.
Calculate the probability of more than 2 claims in a week.
A free 48-minute diagnostic across all three topics. No account. The course has 138 practice questions with every answer worked, and 2 full practice exams.
The syllabus names binomial, geometric, hypergeometric, negative binomial, Poisson and uniform among the discrete distributions, and beta, exponential, gamma, normal and uniform among the continuous ones.
Use the complement: P(X ≥ k) = 1 − P(X ≤ k − 1). That replaces a long or endless sum with the first k terms.
Binomial counts successes in a fixed number of independent trials with the same chance each time. Poisson counts events in a fixed time or space at a steady average rate, with no fixed number of trials.
Given that an exponential wait has already lasted some time, the remaining wait has the same distribution as a fresh one: P(X > s + t | X > s) = P(X > t).
No. Only six Texas Instruments calculators are approved, and none finds these probabilities for you. Use this page to check your working while you practice.