Chapter 4.1, Problem 44EC

Wheel Spinner The wheel spinner shown here is spun twice. Find the sample space, and then determine the probability of the following events.

a. An odd number on the first spin and an even number on the second spin (Note: 0 is considered even.)

b. A sum greater than 4

c. Even numbers on both spins

d. A sum that is odd

e. The same number on both spins

To determine

To find: The sample space:

The probability of an odd number on the first spin and an even number on the second spin.

Answer to Problem 44EC

The sample space is,

$\left\{\begin{array}{l}\left(0,0\right),\left(0,1\right),\left(0,2\right),\left(0,3\right),\left(0,4\right)\\ \left(1,0\right),\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right)\\ \left(2,0\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right)\\ \left(3,0\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right)\\ \left(4,0\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right)\end{array}\right\}$

The probability of an odd number on the first spin and an even number on the second spin is 0.24.

Explanation of Solution

Given info:

The information shows in Wheel spinner and the wheel spinner spun twice. Also 0 is considered as even.

Calculation:

The possibilities for wheel spinner spun twice is,

$\left\{\begin{array}{l}\left(0,0\right),\left(0,1\right),\left(0,2\right),\left(0,3\right),\left(0,4\right)\\ \left(1,0\right),\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right)\\ \left(2,0\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right)\\ \left(3,0\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right)\\ \left(4,0\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right)\end{array}\right\}$

Thus, the total number of outcomes is 25.

Let event A denote that the outcome as odd number on the first spin and an even number on the second spin. Hence, the possible outcomes to get the odd number on the first spin and an even number on the second spin are ‘ $\left\{\left(1,0\right),\left(1,2\right),\left(1,4\right),\left(3,0\right),\left(3,2\right),\left(3,4\right)\right\}$’. That is, there are 6 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 6 for ‘Number of outcomes in A’ and 25 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{6}{\text{25}}\\ =0.24\end{array}$

Thus, the probability of an odd number on the first spin and an even number on the second spin is 0.24.

To determine

The probability of a sum is greater than 4.

Answer to Problem 44EC

The probability of a sum is greater than 4 is 0.4.

Explanation of Solution

Calculation:

Let event A denote that the outcome as sum is greater than 4. Hence, the possible outcomes to get the sum is greater than 4 are,

‘ $\left\{\left(1,4\right),\left(2,3\right),\left(2,4\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right)\right\}$’.

That is, there are 10 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 10 for ‘Number of outcomes in A’ and 25 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{10}{\text{25}}\\ =0.4\end{array}$

Thus, the probability of a sum is greater than 4 is 0.4.

To determine

The probability of even numbers on both spins.

Answer to Problem 44EC

The probability of even numbers on both spins is 0.36.

Explanation of Solution

Calculation:

Let event A denote that the even numbers on both spins. Hence, the possible outcomes to get even numbers on both spins are,

‘ $\left\{\left(0,0\right),\left(0,2\right),\left(0,4\right),\left(2,0\right),\left(2,2\right),\left(2,4\right),\left(4,0\right),\left(4,2\right),\left(4,4\right)\right\}$’.

That is, there are 9 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 9 for ‘Number of outcomes in A’ and 25 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{9}{\text{25}}\\ =0.36\end{array}$

Thus, the probability of even numbers on both spins is 0.36.

To determine

The probability of a sum that is odd.

Answer to Problem 44EC

The probability of a sum that is odd is 0.48.

Explanation of Solution

Calculation:

Let event A denote that a sum that is odd. Hence, the possible outcomes to get a sum that is odd are, ‘ $\left\{\left(0,1\right),\left(0,3\right),\left(1,0\right),\left(1,2\right),\left(1,4\right),\left(2,1\right),\left(2,3\right),\left(3,0\right),\left(3,2\right),\left(3,4\right),\left(4,1\right),\left(4,3\right)\right\}$’.

That is, there are 12 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 12 for ‘Number of outcomes in A’ and 25 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{12}{\text{25}}\\ =0.48\end{array}$

Thus, the probability of a sum that is odd is 0.48.

To determine

The probability of same number on both spins.

Answer to Problem 44EC

The probability of same number on both spins is 0.2.

Explanation of Solution

Calculation:

Let event A denote that same number on both spins. Hence, the possible outcomes to get probability of same number on both spins,

‘ $\left\{\left(0,0\right),\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\right\}$’.

That is, there are 5 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 5 for ‘Number of outcomes in A’ and 25 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(A\right)=\frac{5}{\text{25}}\\ =0.2\end{array}$

Thus, the probability of same number on both spins is 0.2.