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 beat 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 about to go cruising with a mutual friend and are arguing over who gets to sit in the front seat. Roommate A suggests a game of rock-paper-scissors to settle the dispute. Consider the game of rock-paper-scissors to be a chance experiment. How many outcomes are in the sample space of this chance experiment? 6 2 9 7 Assume that both roommates' choices of rock, paper, or scissors are random, independent, and equally likely. Let R_A, R_B, P_A, P_B, S_A, and S_B denote roommate A choosing rock, roommate B choosing rock, roommate A choosing paper, roommate B choosing paper, roommate A choosing scissors, and roommate B choosing scissors, respectively. Additionally, let the simple event consisting of roommate's A and B both selecting rock be denoted parenthetically as (R_A, R_B), and so on for the other possibilities. Define event A as the event that roommate A loses the game and does not get to sit in the front seat. Event A is composed of the following outcomes: A = {(R_A, P_B), (P_A, S_B)} A = {(R_A, R_B), (R_A, P_B), (P_A, S_B), (S_A, R_B)} A = {(R_A, P_B), (P_A, S_B), (S_A, R_B)} A = {(R_A, S_B), (P_A, R_B), (S_A, P_B)} What is the probability of event A? 0.11 0.67 0.33 0.90 Let event C be the event that the game ends in a tie. Event C is composed of the following outcomes: C = {(R_A, R_B), (P_A, P_B), (S_A, S_B)} C = {(R_A, R_B), {R_A, P_B), (R_A, S_B), (P_A, R_B), (P_A, P_B)} C = {(R_A, R_B), {R_A, P_B), (P_A, P_B), (S_A, P_B), (S_A, S_B)} C = {(R_A, R_B), (R_A, P_B), (R_A, S_B), (P_A, R_B), (P_A, P_B), (P_A, S_B), (S_A, R_B), (S_A, S_B)} What is the probability of event C? 0.30 0.75 0.33 0.11

Answer 1

• event A: 3, 0.33
• event C: 1, 0.33

Explanation:

Given a game of "rock-paper-scissors" between roommates A and B, you want to know the outcomes and probability associated with A losing, and the outcomes and probability associated with a tie result.

Outcomes

There are 3 choices for each of A and B, so 3×3 = 9 possible outcomes. If event A is roommate A loses, and event C is a tie result, then the 9 possible outcomes will correspond to the events ...

• R_A, R_B — C . . . . . . . . . A's choice, B's choice, outcome
• R_A, P_B — A
• R–A, S_B — a wins
• P_A, R_B — a wins
• P_A, P_B — C
• P_A, S_B — A
• S_A, R_B — A
• S_A, P_B — a wins
• S_A, S_B — C

You can see that 3 of the 9 outcomes are associated with event A, 3 are associated with event C, and 3 are neither of events A or C (a wins).

Event A

Event A is {(R_A, P_B), (P_A, S_B), (S_A, R_B)}, probability 0.33

Event C

Event C is {(R_A, R_B), (P_A, P_B), (S_A, S_B)}, probability 0.33