Chapter 0.5, Problem 2E

(a)

To determine

Whether the event “sum of 10 or doubles” is mutually exclusive or not mutually exclusive, also find its probability.

Answer to Problem 2E

The event “sum of 10 or doubles” is not mutually exclusive.

Probability of the event is approx. 0.22.

Explanation of Solution

Given information:

Two dice are rolled.

Such that

Event occurs:

“sum of 10 or doubles”

Calculations:

Since the sum would be 10 when both die roll a 5 and also it would be a double (5, 5), the event will not be mutually exclusive.

We know that

When two dice are rolled,

The number of possible outcomes would be 36.

Such that

For the first die, there are 6 possibilities.

Then

For each number on the first die, there are 6 possibilities for the second die.

Now,

For sum of 10,

There are three possible outcomes:

(4, 6), (5, 5), (6, 4)

For doubles,

There are six possible outcomes:

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

For getting 5 on both dice,

There is only one outcome:

(5, 5)

Now,

Probability for getting sum of 10,

  $P\left(\text{sum}\text{of}\text{10}\right)=\frac{3}{36}$

Probability for getting doubles,

  $P\left(\text{doubles}\right)=\frac{6}{36}$

Probability for getting “sum of 10 and doubles”

  $P\left(\text{sum}\text{of}\text{10}\text{and}\text{doubles}\right)=\frac{1}{36}$

Then

Apply general addition rule, for the probability of event “sum of 10 or doubles”:

  $\begin{array}{l}P\left(\text{sum}\text{of}\text{10}\text{or}\text{doubles}\right)=P\left(\text{sum}\text{of}\text{10}\right)+P\left(\text{doubles}\right)-P\left(\text{sum}\text{of}\text{10}\text{and}\text{doubles}\right)\\ =\frac{3}{36}+\frac{6}{36}-\frac{1}{36}\\ =\frac{8}{36}\\ =\frac{2}{9}\\ \approx 0.22\end{array}$

Thus,

The event is not mutually exclusive with the probability of approx. 0.22.

(b)

To determine

Whether the event “sum of 6 or 7” is mutually exclusive or not mutually exclusive, also find its probability.

Answer to Problem 2E

The event“sum of 6 or 7” is mutually exclusive.

Probability of the event is approx. 0.31.

Explanation of Solution

Given information:

Two dice are rolled.

Such that

Event occurs:

“sum of 10 or doubles”

Calculations:

Since both dice cannot have sum of 6 or 7 at the same time, the event will be mutually exclusive.

We know that

When two dice are rolled,

The number of possible outcomes would be 36.

Such that

For the first die, there are 6 possibilities.

Then

For each number on the first die, there are 6 possibilities for the second die.

Now,

For sum of 6,

There are five possible outcomes:

(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)

For sum of 7,

There are six possible outcomes:

(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

Now,

Probability for getting sum of 6,

  $P\left(\text{sum}\text{of}6\right)=\frac{5}{36}$

Probability for getting sum of 7,

  $P\left(\text{sum}\text{of}7\right)=\frac{6}{36}$

Then

Apply general addition rule for mutually exclusive events, for the probability for “sum of 6 or 7”:

  $\begin{array}{l}P\left(\text{sum}\text{of}6\text{or}7\right)=P\left(\text{sum}\text{of}6\right)+P\left(\text{sum}\text{of}\text{7}\right)\\ =\frac{5}{36}+\frac{6}{36}\\ =\frac{11}{36}\\ \approx 0.31\end{array}$

Thus,

The event is mutually exclusive with the probability of approx. 0.31.

(c)

To determine

Whether the event “sum < 3 or sum > 10” is mutually exclusive or not mutually exclusive, also find its probability.

Answer to Problem 2E

The event “sum < 3 or sum > 10” is mutually exclusive.

Probability of the event is approx. 0.11.

Explanation of Solution

Given information:

Two dice are rolled.

Such that

Event occurs:

“sum of 10 or doubles”

Calculations:

Since the sum can be either less than 3 or greater than 10 when both dice are thrown, the event will be mutually exclusive.

For sum less than 3,

There is only one possible outcome:

(1, 1)

For sum greater than 10,

There are three possible outcomes:

(5, 6), (6, 5), (6, 6)

Since there are 4 possible outcomes that satisfy the requirement of getting sum less than 3 or greater than 10

Thus,

The probability for“sum < 3 or sum > 10”:

  $\begin{array}{l}P\left(\text{sum}<3\text{or}\text{sum}>10\right)=\frac{\text{Outcomes}\text{that}\text{satisfy}\text{requirements}}{\text{Total}\text{outcomes}}\\ \text{=}\frac{4}{36}\\ =\frac{1}{9}\\ \approx 0.11\end{array}$

The event is mutually exclusive with the probability of approx. 0.11.