A recent survey conducted by Towers Perrin and published in the Financial Times showed that among 460 organizations in 13 European countries, 93% have bonus plans, 55% have cafeteria-style benefits, and 70% employ home-based workers. If the types of benefits are independent, what is the probability that an organization selected at random will have at least one of the three types of benefits?
Solution
Let $A$ denote organizations have bonus plan, $P(A)=0.93$. So $P(A')=1- 0.93 = 0.07$.
Let $B$ denote organizations have Cafeteria-style benefits, $P(B) =0.55$. So $P(B')=1- 0.55 = 0.45$.
Let $C$ denote organizations have employ home-based workers, $P(C) =0.70$. So $P(C')=1- 0.70 = 0.3$.
The probability that an organization selected at random have at least 1 of the three types of benefits is equal to 1 minus the probability that none of the benefits.
The probability that an organization selected at random will have none of the benefits = $P(A'\cap B' \cap C')$.
Thus
$$ \begin{aligned} P(A'\cap B' \cap C') &= P(A')*P(B')*P(C')\\ &= 0.07 * 0.45 * 0.3\\ &= 0.00945 \end{aligned} $$
The probability that an organization selected at random will have at least 1 of the three types of benefits is
$= 1- P(A'\cap B' \cap C') =1- 0.00945= 0.99055$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
