A consulting engineer receives on average 0.7 requests per week. Find the probability that

a) In a given week, there will be at least one request;

b) In a given 4 week period there will be at least 3 requests.

#### Solution

Given that $X\sim P(0.7)$.

The probability mass function of $X$ is

$$

P(X=x) = \frac{e^{-0.7}0.7^x}{x!},\; x = 0,1,2,\cdots

$$

(a) The probability that in a given week there will be at least one request is

` $$ \begin{aligned} P(X \geq 1) & =1- P(X=0)\\\\ & =1- \frac{e^{-0.7}0.7^0}{0!}\\\\ & = 1- 0.4966\\\\ & = 0.5034 \end{aligned} $$ `

(b) $X\sim P(0.7\times 4)$. That is $X\sim P(2.8)$

The probability mass function of $X$ is

` $$ P(X=x) = \frac{e^{-2.8}2.8^x}{x!},\; x = 0,1,2,\cdots $$ `

The probability that in a given 4 week there will be at least 3 request is

` $$ \begin{aligned} P(X \geq 3) & =1- \sum_{x=0}^{2} P(X=x)\\\\ & =1-\sum_{x=0}^{2} \frac{e^{-2.8}2.8^x}{x!}\\\\ & = 1-[0.0608+0.1703+0.2384]\\\\ & = 1-0.4695\\\\ & = 0.5305\\\\ \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators