Chapter 4.3, Problem 3CYU

a.

To determine

Explain the reason for independence of event A and B.

Explanation of Solution

Two dies are rolled. Event A denotes that number on the first die is even and B denotes that number on the second die is 6.

Independent events:

Two events are said to be independent if occurrence of one does not affect the occurrence of other.

Here, event A and B denotes that number on the first die is even and the number on the second die is 6 respectively. The number on second die does not depend on number on first die. Therefore, events A and B are independent.

Therefore, events A and B are independent.

b.

To determine

Compute the value of $P\left(A\right),P\left(B\right),\text{and }P\left(A\text{and }B\right)$.

Answer to Problem 3CYU

$\begin{array}{l}\underset{\_}{P\left(A\right)=\frac{1}{2}}\\ \underset{\_}{P\left(B\right)=\frac{1}{6}}\\ \underset{\_}{P\left(A\text{and }B\right)=\frac{1}{12}}\end{array}$.

Explanation of Solution

Calculation:

The probability of the event E can be obtained by the formula:

$P\left(E\right)=\frac{\text{Number of outcomes in }E}{\text{Number of outcomes in}\text{sample space}}$

Sample space:

Set of all possible outcomes of probability experiment is called sample space of the experiment.

Here two fair dice are rolled. Outcomes are occurred in ordered pairs. First and second number in each ordered pair represents the number comes up in the first and second die respectively. Each of the two numbers can take values 1 to 6.

Therefore,

$\text{Sample space}=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

For the event A:

Event A denotes that number on the first die is even.

Event A contain those outcomes with first number in the ordered pair are even.

Here, $A=\left\{\begin{array}{l}\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

Among the 36 outcomes, 18 of them are such outcomes.

Substitute 18 for “number of outcomes in A” and 36 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(A\right)=\frac{18}{36}\\ =\frac{1}{2}\\ =0.5\end{array}$

Thus, $P\left(A\right)=\frac{1}{2}$.

For the event B:

Event B denotes that number on the second die is 6.

Event B contain those outcomes with second number in the ordered pair is 6.

Here, $B=\left\{\left(1,6\right),\left(2,6\right),\left(3,6\right),\left(4,6\right),\left(5,6\right),\left(6,6\right)\right\}$

Among the 36 outcomes, 6 of them contain second digit as 6.

Substitute 6 for “number of outcomes in B” and 36 for “Number of outcomes in sample space” in the probability formula.

Therefore,

$\begin{array}{c}P\left(B\right)=\frac{6}{36}\\ =\frac{1}{6}\end{array}$

Thus, $P\left(B\right)=\frac{1}{6}$.

For the event ‘A and B’:

Event ‘A and B’ denote that number on the first die is even and that in second die is 6.

Multiplication Rule for Independent events:

If the events A and B then $P\left(A\text{and }B\right)=P\left(A\right)\times P\left(B\right)$.

Here, $P\left(A\right)=\frac{1}{2}\text{and}P\left(B\right)=\frac{1}{6}$. Substitute these values in the multiplication rule.

Therefore,

$\begin{array}{c}P\left(A\text{and }B\right)=\frac{1}{2}\times \frac{1}{6}\\ =\frac{1}{12}\end{array}$

Thus, $P\left(A\text{and }B\right)=\frac{1}{12}$.