Chapter 9.7, Problem 61E

a.

To determine

To find: the probability of ball landing in the number 00 pocket.

Answer to Problem 61E

  $\frac{1}{38}$

Explanation of Solution

Let E be the event of ball landing in the number 00 pocket.

Since there is only one 00 pocket, so $n\left(E\right)=1$

There are in total 38 pockets, so $n\left(S\right)=38$

So, the probability of ball landing in the number 00 pocket is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{1}{38}\end{array}$

b.

To determine

To find: the probability of ball landing in a red pocket.

Answer to Problem 61E

  $\frac{9}{19}$

Explanation of Solution

Let E be the event of ball landing in a red pocket.

Since there are 18 red pockets, so $n\left(E\right)=18$

There are in total 38 pockets, so $n\left(S\right)=38$

So, the probability of ball landing in a red pocket is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{18}{38}\\ =\frac{9}{19}\end{array}$

c.

To determine

To find: the probability of ball landing in a green pocket or a black pocket.

Answer to Problem 61E

  $\frac{10}{19}$

Explanation of Solution

Let E be the event of ball landing in a green pocket or a black pocket.

Since there are 18 black pockets and 2 green pockets, so $n\left(E\right)=18+2=20$

There are in total 38 pockets, so $n\left(S\right)=38$

So, the probability of ball landing in a green pocket or a black pocket is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{20}{38}\\ =\frac{10}{19}\end{array}$

d.

To determine

To find: the probability of ball landing in the number 14 pocket on two consecutive spins.

Answer to Problem 61E

  $\frac{1}{1444}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event of ball landing in the number 14 pocket on two consecutive spins.

Let A be even of ball landing in number 14 pocket.

Since there is only 1 number 14 pocket, so $n\left(A\right)=1$

There are in total 38 pockets, so $n\left(S\right)=38$

So, the probability of ball landing in number 14 pocket is,

  $\begin{array}{c}P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\\ =\frac{1}{38}\end{array}$

Now, the events of ball landing in any pocket are independent events.

So, the probability of ball landing in the number 14 pocket on two consecutive spins is:

  $\begin{array}{c}P\left(E\right)=\frac{1}{38}\times \frac{1}{38}\\ =\frac{1}{1444}\end{array}$

e.

To determine

To find: the probability of ball landing in a red pocket on three consecutive spins.

Answer to Problem 61E

  $\frac{729}{6859}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event of ball landing in a red pocket on three consecutive spins.

Let A be even of ball landing in red pocket.

From part (b), the probability of ball landing in a red pocket is:

  $P\left(A\right)=\frac{9}{19}$

Now, the events of ball landing in any pocket are independent events.

So, the probability of ball landing in a red pocket on three consecutive spins is:

  $\begin{array}{c}P\left(E\right)=\frac{9}{19}\times \frac{9}{19}\times \frac{9}{19}\\ =\frac{729}{6859}\end{array}$