Chapter 13.2, Problem 17PPS

(a)

To determine

To Find: The probability that in a row of 8 pool balls, the solid 2 and striped 11 would be first and the second from the left.

Answer to Problem 17PPS

The required probability is $\frac{1}{56}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The probability that the 2 ball and the 11 ball will be the first and the second is the row is equal to the number of ways in the remaining 6 balls can be arranged divided by the total number of ways 8 ball pools can be arranged.

Then,

  $\begin{array}{c}P=\frac{6!}{8!}\\ =\frac{1}{56}\end{array}$

(b)

To determine

To Find: Theprobability that if the 8 pool balls were mixed up at the random then they would end up in the given order.

Answer to Problem 17PPS

The required probability is $\frac{1}{40320}$ .

Explanation of Solution

The probability that the randomly selected 8 pool balls would be in the exact order is equal to the singular outcome divided by the total number of ways 8 pool balls are arranged that is,

  $\begin{array}{c}P=\frac{1}{8!}\\ =\frac{1}{40320}\end{array}$

(c)

To determine

To Find: The probability that in the row of the seven ball with the three 8 balls there 9 balls and one 6 balls, the three 8 balls would be to the left of the 6 balls and the 3 9 balls on the right.

Answer to Problem 17PPS

The required probability is $\frac{1}{140}$ .

Explanation of Solution

The probability that in the row of the seven ball with the three 8 balls there 9 balls and one 6 balls, the three 8 balls would be to the left of the 6 balls and the 3 9 balls on the right is,

  $\begin{array}{c}P\left(8,8,8,6,9,9,9\right)=\frac{1}{\frac{7!}{3!3!}}\\ =\frac{1}{140}\end{array}$

(d)

To determine

To Find: The probability that the 6 ball is next to the 7 ball when the ball are arranged randomly.

Answer to Problem 17PPS

The required probability is $\frac{2}{7}$ .

Explanation of Solution

The probability that the 6 ball is next to the 7 ball when the ball are arranged randomly is,

  $\begin{array}{c}P\left(6\text{next}\text{to}7\right)=\frac{2\left(6!\right)}{7!}\\ =\frac{1440}{5040}\\ =\frac{2}{7}\end{array}$