A common way for two people to settle a frivolous dispute is to play a game of rock-paper-scissors. In this game, each person simultaneously displays a hand signal to indicate a rock, a piece of paper, or a pair of scissors. Rock beats scissors, scissors beats paper, and paper beats rock. If both players select the same hand signal, the game results in a tie. Two roommates, roommate A and roommate B, are expecting company and are arguing over who should have to wash the dishes before the company arrives. Roommate A suggests a game of rock-paper-scissors to settle the dispute. Consider the game of rock-paper-scissors to be an experiment. In the long run, roommate A chooses rock 21% of the time, and roommate B chooses rock 61% of the time; roommate A selects paper 39% of the time, and roommate B selects paper 21% of the time; roommate A chooses scissors 40% of the time, and roommate B chooses scissors 18% of the time. (These choices are made randomly and independently of each other.) The probabilities were assigned using the . Define event B as the event that roommate B wins the game and thus does not have to wash the dishes. What is P(B), the probability of event B? P(B) = 0.22 P(B) = 0.36 P(B) = 0.43 P(B) = 0.67 Let event C be the event that the game ends in a tie. What is P(C), the probability of event C? P(C) = 0.55 P(C) = 0.32 P(C) = 0.25 P(C) = 0.28 Define event A as the event that roommate A wins the game and thus does not have to wash the dishes. What is the probability that roommate A wins the game? P(A) = 0.67 P(A) = 0.39 P(A) = 0.37 P(A) = 0.36 What is the complement of event B? BcBc = event B BcBc = event A or event C BcBc = event B or event C BcBc = event A The probability of BcBc is .

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter2: Working With Real Numbers

Section2.7: Problem Solving: Consecutive Integers

Problem 1OE

See similar textbooks

Related questions

Question

A common way for two people to settle a frivolous dispute is to play a game of rock-paper-scissors. In this game, each person simultaneously displays a hand signal to indicate a rock, a piece of paper, or a pair of scissors. Rock beats scissors, scissors beats paper, and paper beats rock. If both players select the same hand signal, the game results in a tie.

Two roommates, roommate A and roommate B, are expecting company and are arguing over who should have to wash the dishes before the company arrives. Roommate A suggests a game of rock-paper-scissors to settle the dispute.

Consider the game of rock-paper-scissors to be an experiment. In the long run, roommate A chooses rock 21% of the time, and roommate B chooses rock 61% of the time; roommate A selects paper 39% of the time, and roommate B selects paper 21% of the time; roommate A chooses scissors 40% of the time, and roommate B chooses scissors 18% of the time. (These choices are made randomly and independently of each other.)

The probabilities were assigned using the .

Define event B as the event that roommate B wins the game and thus does not have to wash the dishes.

What is P(B), the probability of event B?

P(B) = 0.22

P(B) = 0.36

P(B) = 0.43

P(B) = 0.67

Let event C be the event that the game ends in a tie.

What is P(C), the probability of event C?

P(C) = 0.55

P(C) = 0.32

P(C) = 0.25

P(C) = 0.28

Define event A as the event that roommate A wins the game and thus does not have to wash the dishes. What is the probability that roommate A wins the game?

P(A) = 0.67

P(A) = 0.39

P(A) = 0.37

P(A) = 0.36

What is the complement of event B?

BcBc = event B

BcBc = event A or event C

BcBc = event B or event C

BcBc = event A

The probability of BcBc is .

Expert Solution