Chapter 14, Problem 35E

a)

To determine

To find the probability that you roll all 6’s.

Answer to Problem 35E

The probability that you roll all 6’s = 0.0046

Explanation of Solution

Given:

You roll a fair die three times.

Formula:

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

The probability of each outcomes of fair die is,

  $\begin{array}{l}P(X=x)=\frac{1}{6},\\ Wherex=1,2,3,4,5,6\end{array}$

Using multiplication rule,

  $\begin{array}{l}P(3sixes)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\\ P(3sixes)=0.0046\end{array}$

b)

To determine

To find the probability that you roll all odd numbers.

Answer to Problem 35E

The probability that you roll all odd numbers = 0.125

Explanation of Solution

Given:

You roll a fair die three times.

Formula:

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Addition rule:

  $P(AorB)=P(A)+P(B)$

The probability of each outcomes of fair die is,

  $\begin{array}{l}P(X=x)=\frac{1}{6},\\ Wherex=1,2,3,4,5,6\end{array}$

Using addition rule,

  $P(odd)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}=0.5$

Using multiplication rule,

  $\begin{array}{l}P(3odd)=0.5\times 0.5\times 0.5\\ P(3odd)=0.125\end{array}$

Hence, probability that you roll all odd numbers is 0.125.

c)

To determine

To find the probability that none of your roll gets number not divisible by 3.

Answer to Problem 35E

The probability that none of your roll gets number not divisible by 3 = 0.2963

Explanation of Solution

Given:

You roll a fair die three times.

Formula:

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Addition rule:

  $P(AorB)=P(A)+P(B)$

The probability of each outcomes of fair die is,

  $\begin{array}{l}P(X=x)=\frac{1}{6},\\ Wherex=1,2,3,4,5,6\end{array}$

Using addition rule,

  $\begin{array}{l}P(notdivisibleby3)=P(1)+P(2)+P(4)+P(5)\\ P(notdivisibleby3)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{2}{3}\end{array}$

Using multiplication rule,

  $\begin{array}{l}P(3notdivisibleby3)=\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}\\ P(3notdivisibleby3)=0.2963\end{array}$

Hence, probability that none of your roll gets number not divisible by 3 is 0.2963.

d)

To determine

To find the probability that you roll at least one 5.

Answer to Problem 35E

The probability that you roll at least one 5 = 0.4213

Explanation of Solution

Given:

You roll a fair die three times.

Formula:

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Addition rule:

  $P(AorB)=P(A)+P(B)$

The probability of each outcomes of fair die is,

  $\begin{array}{l}P(X=x)=\frac{1}{6},\\ Wherex=1,2,3,4,5,6\end{array}$

Using complement rule,

  $\begin{array}{l}P(not5)=1-P(5)\\ P(not5)=1-\frac{1}{6}=\frac{5}{6}\end{array}$

Using multiplication rule,

  $P(3not5)=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}=\frac{125}{216}$

Therefore, using complement rule,

  $P(atleastone5)=1-\frac{125}{216}=0.4213$

Hence, probability that you roll at one least 5 is 0.4213

e)

To determine

To find the probability that the numbers you roll are not all 5.

Answer to Problem 35E

The probability that the numbers you roll are not all 5 = 0.9954

Explanation of Solution

Given:

You roll a fair die three times.

Formula:

Multiplication rule:

  $P(AandB)=P(A)\times P(B)$

Addition rule:

  $P(AorB)=P(A)+P(B)$

The probability of each outcomes of fair die is,

  $\begin{array}{l}P(X=x)=\frac{1}{6},\\ Wherex=1,2,3,4,5,6\end{array}$

Using multiplication rule,

  $P(3fives)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{216}$

Therefore, using complement rule,

  $P(notall5)=1-\frac{1}{216}=0.9954$

Hence, probability that the numbers you roll are not all 5 is 0.9954