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The article 'should You Report That Fender Bender?" (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.l 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?

Short Answer

Expert verified

a)The probability that at most 4 involve a single vehicle is \(B\left( {4{\rm{ }};{\rm{ }}15,0.7} \right) = 0.001\).

b)The probability that exactly 4 involve a single vehicle is \(b\,(4;15,\,0.7) = 0.001.\)

c)The probability that exactly 6 involve multiple vehicles is \(\;b\,\left( {6{\rm{ }};{\rm{ }}15,0.3} \right) = 0.147\).

d)The probability that between 2 and 4 , inclusive, involve a single vehicle is P(2\( \le X \le 4) = 0.001\).

e)The probability that at least 2 involve a single vehicle is \(P(2 \le X) = 1\).

f) The probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles must be \(0.001\).

Step by step solution

01

Definition of appendix table

The appendix of a paper contains research-related information that isn't necessary to include in the main body.

02

Step 2: What is the probability that at most 4 involve a single vehicle

a)
The cdf of a binomial random variable X with parameters n and p is the Cumulative Density Function.

\(\begin{aligned}B(x;n,p) = P(X \le x)\\ &= \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n.\end{aligned}\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{\rm{ }}\end{array}} \right.\)

(a):

Using the codf and theorem given above, for \(n = 15,x = 4\) (at most 4), and \(p = 0.7\) ( 7 in 10 auto accidents invole a single vehicle), the following is true

\(\begin{aligned}B(4;15,0.7) = P(X \le 4)\\ &= \sum\limits_{y = 0}^4 b (y;15,0.7)\mathop = \limits^{(1)} 0.001\end{aligned}\)

(1): from the table of Appendix Table A.l.

Therefore,\(B\left( {4{\rm{ }};{\rm{ }}15,0.7} \right) = 0.001\).

03

Step 3: What is the probability that exactly 4 involve a single vehicle

b)

The cdf of a binomial random variable X with parameters n and p is a Cumulative Density Function.

\(\begin{aligned}B(x;n,p) = P(X \le x)\\ &= \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\end{aligned}\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{\rm{ }}\end{array}} \right.\)

(b):

Using the codf and theorem given above, for\(n = 15,x = 4\), and\(p = 0.7\)( 7 in 10 auto accidents involve a single vehicle), the following is true

\(\begin{aligned}b(4;15,0.7)\mathop = \limits^{(1)} B(4;15,0.7) - B{(3;15,0.7)^{(2)}}\\ &= 0.001 - 0.000\\ &= 0.001\end{aligned}\)

(1): We're expected to use Appendix Table A.l., where the only method to get exactly four that involve a single vehicle is to deduct at most four from at most three.

(2): from the Appendix Table A.l.

Therefore, \(b\,(4;15,\,0.7) = 0.001.\)

04

Step 4: What is the probability that exactly 6 involve multiple vehicles

c)

The cdf of a binomial random variable X with parameters n and p is a Cumulative Density Function.

\(\begin{array}{l}B(x;n,p) = P(X \le x)\\ = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\end{array}\)

Theorem:

\(b(x;n,p) = \left\{ {\begin{array}{*{20}{l}}{\left( {\begin{array}{*{20}{l}}n\\x\end{array}} \right){p^x}{{(1 - p)}^{n - x}}}&{,x = 0,1,2, \ldots ,n}\\0&{\rm{ }}\end{array}} \right.\)

(c):

Using the cdf and theorem given above, for \(n = 15,x = 6\), and \(p = 0.3\) ( 3 in 10 auto accidents involve a multiple vehicles), the following is true

\(\begin{aligned}b(6;15,0.3)\mathop = \limits^{(1)} B(6;15,0.3) - B{(6;15,0.3)^{(2)}}\\ &= 0.869 - 0.722\\ &= 0.147\end{aligned}\)

(1) : We're expected to use Appendix Table A.l., where the only method to get exactly four that involve a single vehicle is to deduct at most four from at most three.

(2): from the Appendix Table A.l.

Therefore, \(\;b\,\left( {6{\rm{ }};{\rm{ }}15,0.3} \right) = 0.147\).

05

Step 5: What is the probability that between 2 and 4 , inclusive, involve a single vehicle

d)

Cumulative Density Function cdf of binomial random variable X with parameters n and p is

\(\begin{array}{c}B(x;n,p) = P(X \le x)\\ = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n.\end{array}\)

(d):

For\(n = 15\), and\(p = 0.7\)( 7 in 10 auto accidents involve a single vehicle), the following is true

\(\begin{aligned}P(2 \le X \le 4)\mathop = \limits^{(1)} B(4;15,0.7) - B{(1;15,0.7)^{(2)}}\\ &= 0.001 - 0.000\\ &= 0.001\end{aligned}\)

(1) : we are supposed to use Appendix Table A.l. where the only way to do is to subtract at most four with at most 1 to get 2 , 3 or 4 single vehicle accidents;

(2) : from the Appendix Table A.l.

Therefore, \(P(2 \le X \le 4) = 0.001\).

06

Step 6: What is the probability that at least 2 involve a single vehicle

e)

Cumulative Density Function cdf of binomial random variable X with parameters n and p is

\(\begin{array}{c}B(x;n,p) = P(X \le x)\\ = \sum\limits_{y = 0}^x b (y;n,p),\;\;\;x = 0,1, \ldots ,n\end{array}\)

(e):

For\(n = 15\), and\(p = 0.7\)( 7 in 10 auto accidents involve a single vehicle), the following is true

\(\begin{aligned}P(2 \le X)\mathop = \limits^{(1)} 1 - P(X \le 1) = 1 - B{(1;15,0.7)^{(2)}}\\ &= 1 - 0.000\\ &= 1\end{aligned}\)

(1): we are supposed to use Appendix Table A.l, this is one of the reasons we use complement (for any event A, \(P\left( {{A^\prime }} \right) = 1 - P(A)\));

(2) : from the Appendix Table A.l.

Therefore, P(2 \leq X)=1.

07

Step 7: What is the probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles

f)

Because four of them are single, the others must be many. The answer must be \(0.001\), which is the same as b).

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