A common way for two people to settle a frivolous dispute is to play a game of rock-paper-scissors. In this game, each p

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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 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 62% of the time, and roommate
B chooses rock 62% of the time; roommate A selects paper 16% of the
time, and roommate B selects paper 23% of the time; roommate A
chooses scissors 22% of the time, and roommate B chooses scissors
15% of the time. (These choices are made randomly and independently
of each other.)
The probabilities were assigned using the (classical
approach-relative frequency approach-subjective
approach)
What is P(B), the probability of event B?
P(B) = 0.67
P(B) = 0.22
P(B) = 0.30
P(B) = 0.76
Let event C be the event that the game ends in a tie.
What is P(C), the probability of event C?
P(C) = 0.45
P(C) = 0.25
P(C) = 0.86
P(C) = 0.32
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) = 0.39
P(A) = 0.50
P(A) = 0.24
P(A) = 0.67
What is the complement of event B?
BcBc = event A
BcBc = event A or event C
BcBc = event C
BcBc = event B or event C
The probability of BcBc is
(0.24-0.35-0.75-0.70)