A researcher wishes to conduct a study of the color preferences of new car buyers. Suppose that 30% of this population prefers the color brown. If 12 buyers are randomly selected, what is the probability that exactly 5 buyers would prefer brown?
Solution
Let $X$ denote the number of buyers who prefer brown color out of 12 randomly selected buyers.
Let $p$ denote the probability that buyer prefer brown color.
Given that $n= 12$ and $p=0.3$.
The random variable $X$ follows Binomial distribution. That is $X\sim B(n=12, p=0.3)$.
The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{12}{x} (0.3)^x (1-0.3)^{12-x}, \\ &\quad x=0,1,\cdots, 12. \end{aligned} $$
The probability that exactly 5 buyers would prefer brown is
$$ \begin{aligned} P(X= 5) & =\binom{12}{5} (0.3)^{5} (1-0.3)^{12-5}\\ & = 0.1585\\ \end{aligned} $$
Thus the probability that exactly 5 buyers would prefer brown is $0.1585$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
