Of the 28 professors in a certain department, 18 drive foreign and 10 drive domestic cars. If five of these professors are selected at random, what is the probability that at least three of them drive foreign cars?
Solution
The probability that a randomly seleceted professor drive foreign car is $p = \frac{18}{28}=0.6429$.
$n = 5$ professors are selected at random.
The random variable $X$ is the number of professors who drive foreign car is $X=0,1,2,3,4,5$.
So $X\sim B(5, 0.6429)$ distribution.
The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{5}{x} (0.6429)^x (1-0.6429)^{5-x},\\ &\quad x=0,1,\cdots, 5. \end{aligned} $$
The probability that at least three professor drive foreign car is $P(X\geq 3)$.
Thus
$$ \begin{aligned} P(X\geq 3) & =\sum_{x=3}^{5} P(x)\\ & =\sum_{x=3}^{5}\binom{5}{x}(0.6429)^x(1-0.6429)^{5-x}\\ & = (0.3389)+(0.305)+(0.1098) \\ & = 0.7537 \end{aligned} $$
Hence the probability that at least three of them drive foreign car is $0.7537$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
