Chapter 5.2, Problem 39E

(a)

To determine

To explain:

The sample space for role playing games for the die twice

Explanation of Solution

Given:

On the first roll is possible to roll a $1,2,3\text{and}4$

Concept used:

Sample space:-In probability theory the sample space of an experiment or random trial is the set of all possible outcomes or result of that experiment

Calculation:

A sample space is usually denoted using set notation and the possible ordered outcomes are listed as element in the set

On the first roll is possible to roll a $1,2,3\text{and}4$

On the second roll it is possible to roll a $1,2,3\text{and}4$

The sample sequence then contains all possible combinations of outcomes on the first and second roll

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

(b)

To determine

To find:

The assignment of probability to outcome in this sample space

Answer to Problem 39E

The probability $=\frac{1}{16}$

Explanation of Solution

Given:

On the first roll is possible to roll a $1,2,3\text{and}4$

Concept used:

Sample space:-In probability theory the sample space of an experiment or random trial is the set of all possible outcomes or result of that experiment

Calculation:

The sample space of a sequence of die flips is all possible sequences of outcomes$\begin{array}{l}=2^4\\ =16\end{array}$

That is

The sample of a die flip is $\left\{1,2,3,4\right\}$

Since the given sample space has $16$ outcomes

Therefore , the probability of all the event mentioned in the given sample space is same

Hence ,

The probability $=\frac{1}{16}$