Chapter 2, Problem 168SE

As in Exercises 2.166 and 2.167, eight tires of different brands are ranked from 1 to 8 (best to worst) according to mileage performance.

1. a If four of these tires are chosen at random by a customer, what is the probability that the best tire selected is ranked 3 and the worst is ranked 7?
2. b In part (a) you computed the probability that the best tire selected is ranked 3 and the worst is ranked 7. If that is the case, the range of the ranks, R = largest rank − smallest rank = 7 − 3 = 4. What is P(R = 4)?
3. c Give all possible values for R and the probabilities associated with all of these values.

2.167 Refer to Exercise 2.166. Let Y denote the actual quality rank of the best tire selected by the customer. In Exercise 2.166, you computed P(Y = 3). Give the possible values of Y and the probabilities associated with all of these values.

2.166 Eight tires of different brands are ranked from 1 to 8 (best to worst) according to mileage performance. If four of these tires are chosen at random by a customer, find the probability that the best tire among those selected by the customer is actually ranked third among the original eight.

To determine

Find the probability that the best tire selected is ranked 3 and the worst is ranked 7.

Answer to Problem 168SE

The best tire selected is ranked 3 and the worst is ranked 7 is $\frac{3}{70}$.

Explanation of Solution

Calculation:

Combination rule:

If the number of unordered subsets of size r selected from n objects then it can be termed as combination.

The combination formula is shown below.

$\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

Here, n represents the total number of objects and r represents the number of unordered subsets.

The number of ways of selecting four tires from 8 tires is obtained as follows:

$\begin{array}{c}\left(\begin{array}{l}8\\ 4\end{array}\right)=\frac{8!}{4!\left(8-4\right)!}\\ =\frac{8\times 7\times 5\times 4!}{4!\times 4\times 3\times 2\times 1}\\ =70\end{array}$

In this case, the possible ranks to choose other two tires 4, 5 or 6.

The number of ways of selecting 2 tires from 3 tires is obtained as follows:

$\begin{array}{c}\left(\begin{array}{l}3\\ 2\end{array}\right)=\left(\frac{3!}{2!\left(3-2\right)!}\right)\\ =3\end{array}$.

The probability that the best tire selected is ranked 3 and the worst is ranked 7 is obtained as follows:

$\begin{array}{c}\text{Required probability}=\frac{\left(\begin{array}{l}8\\ 4\end{array}\right)}{\left(\begin{array}{l}3\\ 2\end{array}\right)}\\ =\frac{3}{70}\end{array}$

Thus, the probability that the best tire selected is ranked 3 and the worst is ranked 7 is $\frac{3}{70}$.

To determine

Find the probability $P\left(R=4\right)$.

Answer to Problem 168SE

The probability $P\left(R=4\right)$ is $\frac{12}{70}$.

Explanation of Solution

Calculation:

From the given information, the range of the ranks is R = 4.

The possible ways of form range of ranks are {Rank 1 to Rank 5, Rank 2 to Rank 6, Rank 3 to Rank 7, Rank 4 to Rank 8}.

From Part (a), the probability that the best tire selected is ranked 3 and the worst is ranked 7 is $\frac{3}{70}$.

Therefore, each range has probability $\frac{3}{70}$.

The probability $P\left(R=4\right)$ is obtained as follows:

$\begin{array}{c}P\left(R=4\right)=\left(\begin{array}{l}P\left(\text{Rank 1 to Rank 5}\right)+P\left(\text{Rank 2 to Rank 6}\right)+\\ P\left(\text{Rank 3 to Rank 7}\right)+P\left(\text{Rank 4 to Rank 8}\right)\end{array}\right)\\ =\frac{3}{70}+\frac{3}{70}+\frac{3}{70}+\frac{3}{70}\\ =\frac{12}{70}\end{array}$.

Thus, the probability $P\left(R=4\right)$ is $\frac{12}{70}$.

To determine

Find all possible values for R and also find the probabilities associated with all of the R values.

Explanation of Solution

Calculation:

From the given information, the possible values of R are 3, 5, 6 and 7.

The possible ways of forming range (3) of ranks are {Rank 1 to Rank 4, Rank 2 to Rank 5, Rank 3 to Rank 6, Rank 4 to Rank 7, Rank 5 to Rank 8}.

The probability $P\left(R=3\right)$ is obtained as follows:

$\begin{array}{c}P\left(R=3\right)=\frac{\text{Number of ways for getting range 3}}{\text{Total number of cases}}\\ =\frac{5}{70}\\ =0.0714\end{array}$.

Thus, the probability $P\left(R=3\right)$ is 0.0714.

Similarly, the possible ways of getting rank 5, 6, and 7 are 18, 20 and 15.

The probability $P\left(R=5\right)$ is obtained as follows:

$\begin{array}{c}P\left(R=5\right)=\frac{\text{Number of ways for getting range 5}}{\text{Total number of cases}}\\ =\frac{18}{70}\\ =0.2571\end{array}$.

Thus, the probability $P\left(R=5\right)$ is 0.2571.

The probability $P\left(R=6\right)$ is obtained as follows:

$\begin{array}{c}P\left(R=6\right)=\frac{\text{Number of ways for getting range 6}}{\text{Total number of cases}}\\ =\frac{20}{70}\\ =0.2857\end{array}$.

Thus, the probability $P\left(R=6\right)$ is 0.2857.

The probability $P\left(R=7\right)$ is obtained as follows:

$\begin{array}{c}P\left(R=7\right)=\frac{\text{Number of ways for getting range 7}}{\text{Total number of cases}}\\ =\frac{15}{70}\\ =0.2143\end{array}$.

Thus, the probability $P\left(R=7\right)$ is 0.2413.