Chapter 6.3, Problem 37E

Suppose that a six-sided die is weighted so that any even-numbered face is twice as likely to land face up as any odd-numbered face. Consider the chance experiment that consists of rolling this die.

1. a. What are the probabilities of the six simple events? (Hint: Denote these events by O₁, …, O₆. Then P(O₁) = p, P(O₂) = 2p, P(O₃) = p, …, P(O₆) = 2p. Now use a condition on the sum of these probabilities to determine p.)
2. b. What is the probability that the number showing is an odd number? What is the probability that the number showing is at most three?
3. c. Now suppose that the die is weighted so that the probability of each simple event is proportional to the number showing on the corresponding upturned face; that is, P(O₁) = c, P(O₂) = 2c, …, P(O₆) = 6c. What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

To determine

Calculate the probabilities of the simple events.

Answer to Problem 37E

The probabilities of the simple events are given by:

$\begin{array}{l}P\left(O_1\right)=P\left(O_3\right)=P\left(O_5\right)=\frac{1}{9}\\ P\left(O_2\right)=P\left(O_4\right)=P\left(O_6\right)=\frac{2}{9}\end{array}$ .

Explanation of Solution

Calculation:

It is given that even-numbered face is twice as likely to occur as odd-numbered face.

Events $O_i,i=1,2,3,4,5,6$ represent i^th face of a die. It is known that the probability of all sample point in the sample space is 1. Let p be the probability of the odd-numbered face and 2p is the probability of the even-numbered face. The total probability of the six sides is 1.

$\begin{array}{c}\text{Total probability}=1\\ p+2p+p+2p+p+2p=1\\ 9p=1\\ p=\frac{1}{9}\end{array}$

Therefore, the probability of sample events is given by:

$\begin{array}{c}P\left(O_1\right)=P\left(O_3\right)\\ =P\left(O_5\right)\\ =p\\ =\frac{1}{9}\end{array}$

$\begin{array}{c}P\left(O_2\right)=P\left(O_4\right)\\ =P\left(O_6\right)\\ =2p\\ =\frac{2}{9}\end{array}$

Thus, the probabilities of the sample events are given by:

$\begin{array}{l}P\left(O_1\right)=P\left(O_3\right)=P\left(O_5\right)=\frac{1}{9}\\ P\left(O_2\right)=P\left(O_4\right)=P\left(O_6\right)=\frac{2}{9}\end{array}$ .

To determine

Calculate the probability of getting an odd number when a die is rolled.

Calculate the probability that number showing on the die is at most three.

Answer to Problem 37E

The probability of getting an odd number when a die is rolled is $\frac{1}{3}$ .

The probability that number showing on the die is at most three is $\frac{4}{9}$ .

Explanation of Solution

Calculation:

The required probability is given by:

$\begin{array}{l}P\left(\text{getting an odd number}\right)=P\left(O_1\right)+P\left(O_3\right)+P\left(O_5\right)\\ P\left(\text{at most three}\right)=P\left(O_1\right)+P\left(O_2\right)+P\left(O_3\right)\end{array}$

From part (a), it is clear that $P\left(O_1\right)=P\left(O_3\right)=P\left(O_5\right)=\frac{1}{9},P\left(O_2\right)=\frac{2}{9}$ . Substitute these values in the above equation.

Therefore, the required probability is given by:

$\begin{array}{c}P\left(\text{getting an odd number}\right)=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\\ =\frac{3}{9}\\ =\frac{1}{3}\end{array}$.

Thus, the probability of getting an odd number when a die is rolled is $\frac{1}{3}$ .

Therefore, the required probability is given by:

$\begin{array}{c}P\left(\text{at most three}\right)=\frac{1}{9}+\frac{2}{9}+\frac{1}{9}\\ =\frac{4}{9}\end{array}$.

Thus, the probability that number showing on the die is at most three is $\frac{4}{9}$ .

To determine

Calculate the probabilities of the simple events.

Calculate the probabilities of part (b).

Answer to Problem 37E

The probabilities of the simple events are given by:

$P\left(O_1\right)=\frac{1}{23},P\left(O_2\right)=\frac{2}{23},P\left(O_3\right)=\frac{3}{23},P\left(O_4\right)=\frac{4}{23},P\left(O_5\right)=\frac{5}{23},P\left(O_6\right)=\frac{6}{23}$ .

The probability of getting an odd number when a die is rolled is $\frac{9}{21}$ .

The probability that number showing on the die is at most three is $\frac{6}{21}$ .

Explanation of Solution

Calculation:

It is given that even-numbered face is twice as likely to occur as odd-numbered face.

Events $O_i,i=1,2,3,4,5,6$ represent i^th face of a die. It is known that the probability of all sample point in the sample space is 1. Let c be the probability of the face 1 and 2c is the probability of the face ‘2’, and so on. The total probability of the six sides is 1.

$\begin{array}{c}\text{Total probability}=1\\ c+2c+3c+4c+5c+6c=1\\ 21c=1\\ c=\frac{1}{21}\end{array}$

Therefore, the probabilities of sample events are given by:

$\begin{array}{c}P\left(O_1\right)=\frac{1}{21}\\ P\left(O_2\right)=\frac{2}{21}\\ P\left(O_3\right)=\frac{3}{21}\\ P\left(O_4\right)=\frac{4}{21}\\ P\left(O_5\right)=\frac{5}{21}\\ P\left(O_6\right)=\frac{6}{21}\end{array}$

Thus, the probabilities of the simple events are given by:

$P\left(O_1\right)=\frac{1}{23},P\left(O_2\right)=\frac{2}{23},P\left(O_3\right)=\frac{3}{23},P\left(O_4\right)=\frac{4}{23},P\left(O_5\right)=\frac{5}{23},P\left(O_6\right)=\frac{6}{23}$ .

The required probabilities are obtained as follows:

$\begin{array}{c}P\left(\text{getting an odd number}\right)=P\left(O_1\right)+P\left(O_3\right)+P\left(O_5\right)\\ =\frac{1}{21}+\frac{3}{21}+\frac{5}{21}\\ =\frac{9}{21}\end{array}$

Thus, the probability of getting an odd number when a die is rolled is $\frac{9}{21}$ .

$\begin{array}{c}P\left(\text{at most three}\right)=P\left(O_1\right)+P\left(O_2\right)+P\left(O_3\right)\\ =\frac{1}{21}+\frac{2}{21}+\frac{3}{21}\\ =\frac{6}{21}\end{array}$

Thus, the probability that number showing on the die is at most three is $\frac{6}{21}$ .