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