The article "Should You Report That FenderBender?" (Consumer Reports, Sept. 2013: 15) reported that 7 in 10 auto accidents involve a single vehicle (the article recommended always reporting to the insurance company an accident involving multiple vehicles). Suppose 15 accidents are randomly selected. Use Appendix Table A.1 to answer each of the following questions.

a. What is the probability that at most 4 involve a single vehicle?
b. What is the probability that exactly 4 involve a single vehicle?
c. What is the probability that exactly 6 involve multiple vehicles?
d. What is the probability that between 2 and 4, inclusive, involve a single vehicle?
e. What is the probability that at least 2 involve a single vehicle?
f. What is the probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles?

Solution

Here $X$ denote the number of auto accidents that involve single vehicles.

$p$ be the probability that auto accidents involve single vehicles and $q=1-p$ be the probability that auto accidents involve multiple vehicles.

Given that $p=7/10 =0.7$ and $n =15$. Thus$X\sim B(15, 0.7)$.

The probability mass function of $X$ is

$$
P(X=x) = \binom{15}{x} (0.7)^x (1-0.7)^{15-x}, \; x=0,1,\cdots, 15.
$$

(a) The probability that $X$ is at the most $4$ is

$$ \begin{aligned} P(X\leq 4) & =\sum_{x=0}^{4} P(x)\\ & =\sum_{x=0}^{4}\binom{15}{x}(0.7)^x(1-0.7)^{15-x}\\ & = 0.0007 \end{aligned} $$

(b) The probability that $X$ is exactly $4$ is

$$ \begin{aligned} P(X= 4) & =\sum_{x=0}^{4} P(x)-\sum_{x=0}^{3} P(x)\\\\ & = 0.0006\\ \end{aligned} $$

(c) The probability that exactly 6 involve multiple vehicles (i.e. $x=9$ involve single vehicles) is
$$ \begin{aligned} P(X= 9) & =\sum_{x=0}^{9} P(x)-\sum_{x=0}^{8} P(x)\\\\ & = 0.2784 -0.1311\\ & = 0.1472 \end{aligned} $$

(d) The probability that between 2 and 4, inclusive, involve a single vehicle is

$$ \begin{aligned} P(2\leq X\leq 4) & =\sum_{x=0}^{4} P(x)-\sum_{x=0}^{1} P(x)\\ & = 0.0007-0\\ & = 0.0007\\ \end{aligned} $$

(e) The probability that at least $2$ involve single vehicle is

$$ \begin{aligned} P(X\geq 2) & =1-P(X\leq 1)\\ &= 1-\sum_{x=0}^{1} P(x)\\ & = 1-0 \\ & = 1 \\ \end{aligned} $$

(f) The probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles is same as $P(X=4)$.

$$ \begin{aligned} P(X= 4) & =\sum_{x=0}^{4} P(x)-\sum_{x=0}^{3} P(x)\\ & = 0.0006\\ \end{aligned} $$