OPENINTRO:STATISTICS
OPENINTRO:STATISTICS
4th Edition
ISBN: 9781943450077
Author: OPENINTRO
Publisher: amazon.com
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Chapter 3.1, Problem 19GP

a.

To determine

Explain how to represent event Ac.

a.

Expert Solution
Check Mark

Answer to Problem 19GP

Event Ac can be represented as the total that is equal to 12.

Explanation of Solution

Calculation:

The possible outcomes when rolling two dice is given below:

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

The sum of the outcomes is given below:

S={2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12}

Complement of an event:

The complement of event W is the event that W does not occur. That is, Wc represents that all the outcomes are not in W.

Event A represents the outcomes of getting a total that is less than 12 when rolling two dice.

Therefore, the complement of A (Ac) represents that the total is equal to 12.

b.

To determine

Find the probability of P(Ac).

b.

Expert Solution
Check Mark

Answer to Problem 19GP

The probability of P(Ac) is 136.

Explanation of Solution

Calculation:

From Figure 3.5, it can be observed that the total number of favorable outcomes for event Ac is 1.

The total number of outcomes is 36.

The required probability is obtained as follows:

P(Ac)=Favorable number of outcomes for event AcTotal number of outcomes=136

Thus, the probability of P(Ac) is 136 .

c.

To determine

Find the probability of P(A).

c.

Expert Solution
Check Mark

Answer to Problem 19GP

The probability of P(A) is 136.

Explanation of Solution

Calculation:

The total number of outcomes is 36.

The required probability is obtained as follows:

P(A)=1P(Ac)=1136   From Part (b)=3536

Thus, the probability of P(A) is 136.

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Chapter 3 Solutions

OPENINTRO:STATISTICS

Ch. 3.1 - Prob. 16GPCh. 3.1 - Prob. 17GPCh. 3.1 - Prob. 18GPCh. 3.1 - Prob. 19GPCh. 3.1 - Prob. 20GPCh. 3.1 - Prob. 22GPCh. 3.1 - Prob. 23GPCh. 3.1 - Prob. 24GPCh. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10ECh. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.2 - Prob. 28GPCh. 3.2 - Prob. 29GPCh. 3.2 - Prob. 30GPCh. 3.2 - Prob. 31GPCh. 3.2 - Prob. 32GPCh. 3.2 - Prob. 33GPCh. 3.2 - Prob. 35GPCh. 3.2 - Prob. 36GPCh. 3.2 - Prob. 37GPCh. 3.2 - Prob. 38GPCh. 3.2 - Prob. 39GPCh. 3.2 - Prob. 41GPCh. 3.2 - Prob. 43GPCh. 3.2 - Prob. 45GPCh. 3.2 - Prob. 46GPCh. 3.2 - Prob. 13ECh. 3.2 - Prob. 14ECh. 3.2 - Prob. 15ECh. 3.2 - Prob. 16ECh. 3.2 - Prob. 17ECh. 3.2 - Prob. 18ECh. 3.2 - Prob. 19ECh. 3.2 - Prob. 20ECh. 3.2 - Prob. 21ECh. 3.2 - Prob. 22ECh. 3.3 - Prob. 49GPCh. 3.3 - Prob. 51GPCh. 3.3 - Prob. 52GPCh. 3.3 - Prob. 53GPCh. 3.3 - Prob. 23ECh. 3.3 - Prob. 24ECh. 3.3 - Prob. 25ECh. 3.3 - Prob. 26ECh. 3.3 - Prob. 27ECh. 3.3 - Prob. 28ECh. 3.4 - Prob. 55GPCh. 3.4 - Prob. 59GPCh. 3.4 - Prob. 62GPCh. 3.4 - Prob. 63GPCh. 3.4 - Prob. 64GPCh. 3.4 - Prob. 66GPCh. 3.4 - Prob. 67GPCh. 3.4 - Prob. 69GPCh. 3.4 - Prob. 70GPCh. 3.4 - Prob. 29ECh. 3.4 - Prob. 30ECh. 3.4 - Prob. 31ECh. 3.4 - Prob. 32ECh. 3.4 - Prob. 33ECh. 3.4 - Prob. 34ECh. 3.4 - Prob. 35ECh. 3.4 - Prob. 36ECh. 3.5 - Prob. 73GPCh. 3.5 - Prob. 75GPCh. 3.5 - Prob. 37ECh. 3.5 - Prob. 38ECh. 3 - Prob. 39CECh. 3 - Prob. 40CECh. 3 - Prob. 41CECh. 3 - Prob. 42CECh. 3 - Prob. 43CECh. 3 - Prob. 44CECh. 3 - Prob. 45CECh. 3 - Prob. 46CECh. 3 - Prob. 47CE
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