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

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

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