1.83 The probabilities that a service station will pump gas into 0, 1, 2, 3, 4, or 5 or more cars during a certain 30-minute period are 0.03, 0.18, 0.24, 0.28, 0.10, and 0.17, respectively. Find the probability that in this 30-minute period

(a) more than 2 cars receive gas;

(b) at most 4 cars receive gas;

(c) 4 or more cars receive gas.

### Solution

x | 0 | 1 | 2 | 3 | 4 | 5 or more |
---|---|---|---|---|---|---|

P(x) | 0.03 | 0.18 | 0.24 | 0.28 | 0.10 | 0.17 |

(a) The probability that in this 30-minute period more than 2 cars receive gas is

` $$ \begin{aligned} P(X > 2) &= P(X=3) + P(X=4)+ P(X=5)\\ &= 0.28+0.10+ 0.17\\ & = 0.55 \end{aligned} $$ `

(b) The probability that in this 30-minute period at most 4 cars receive gas is

` $$ \begin{aligned} P(X\leq 4) &= 1- P(X > 4)\\ & = 1- 0.17\\ & = 0.83 \end{aligned} $$ `

(c) The probability that in this 30-minute period 4 or more cars receive gas is

` $$ \begin{aligned} P(X\geq 4) &= P(X=4) + P(X=5)\\ & =0.10+ 0.17\\ & = 0.27 \end{aligned} $$ `

#### Further Reading

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