Explanation Definition: If an experiment has n mutually exclusive and equally likely outcomes, and if s is the favourable outcomes, then the probability of any event E denoted as P ( E ) is defined as, P ( E ) = s n = Number of outcomes favorable to E Number of all possible outcomes . Multiplication law of probability: If E and F are independent events, then P ( E ∩ F ) = P ( E ) ⋅ P ( F ) Calculation: Consider the provided set of octahedral dice. The opponent chooses die D . As the opponent chooses die D , we can select either die A or B or C . Consider die A . If die A is chosen and suppose 4 is rolled on die A and 1 is rolled on die D , the game can be won by us. Define A 4 to represent rolling of 4 on die A and D 1 to represent rolling of 1 on die D . Because the rolling of one die is independent of rolling of other die, use multiplication rule of probability. So, the probability of rolling 4 on die A and 1 on die D is: P ( A 4 ∩ D 1 ) = P ( A 4 ) ⋅ P ( D 1 ) = 6 8 ⋅ 4 8 = 3 8 If 0 is rolled on die A , the game cannot be won by us because the die D has numbers 5 and 1 and 0 is less than both the numbers. So, 0 cannot be rolled. Hence, the probability of choosing die A and winning the game is 3 8 . Now, consider die B . Suppose if 3 is rolled on die B , 1 is rolled on die D , the game can be won by us. Define B 3 to represent rolling of 3 on die B and D 1 to represent rolling of 1 on die D . Because the rolling of one die is independent of rolling of other die, use multiplication rule of probability. So, the probability of rolling 3 on die B and 1 on die D is: P ( B 3 ∩ D 1 ) = P ( B 3 ) ⋅ P ( D 1 ) = 8 8 ⋅ 4 8 = 4 8 If 3 is rolled on die B and 5 is rolled on die D , the game cannot be won by us because 3 is less than 5. Hence, the probability of choosing die B and winning the game is 4 8 . Now, finally consider die C . Suppose if 2 is rolled on die C , 1 is rolled on die D , the game can be won by us. Define C2 to represent rolling of 2 on die C and D 1 to represent rolling of 1 on die D . Because the rolling of one die is independent of rolling of other die, use multiplication rule of probability. So, the probability of rolling 2 on die C and 1 on die D is: P ( C 2 ∩ D 1 ) = P ( C 2 ) ⋅ P ( D 1 ) = 6 8 ⋅ 4 8 = 3 8 If 2 is rolled on die C and 5 is rolled on die D , the game cannot be won by us because 2 is less than 5

13th Edition

ISBN: 9781305855588

Author: SMITH

Publisher: CENGAGE CO

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Chapter 13.CR, Problem 18CR

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The octahedral die that would be chosen and the reason for choosing the die if the player rolls his or her chosen die once and the higher number on the top is the winner.

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Chapter 13.CR, Problem 17CR

Chapter 13.CR, Problem 19CR

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The octahedral die that would be chosen and the reason for choosing the die if the player rolls his or her chosen die once and the higher number on the top is the winner.

Solution Summary: The author explains that the octahedral die that would be chosen is die C because it has the highest probability of winning.

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To find: The octahedral die that would be chosen and the reason for choosing the die if the player rolls his or her chosen die once and the higher number on the top is the winner.

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Chapter 13 Solutions

To find:

The octahedral die that would be chosen and the reason for choosing the die if the player rolls his or her chosen die once and the higher number on the top is the winner.