3. On a given evening, two roommates, roommate A and roommate B, are arguing over who should control the television remote. Roommate A suggests a game of rock-paper-scissors to settle the dispute.
It's the first time that the roommates have played each other in rock-paper-scissors, so there is no history of the choices made by either roommate. However, roommate B thinks that the three possibilities for his roommate's choice of hand signals are not equally likely. Based on his roommate's behavior in other similar games, roommate B believes that there is a .56 probability that his roommate will choose rock and a .33 probability that his roommate will choose scissors.
The probabilities are assigned using the
a. empirical method
b. subjective method
c. theoretical method
I think it is B, subjective
Suppose that roommate B's probability assignments are correct. What can you say about P(Pa), the probability that roommate A chooses paper? Check all that apply
a. P(Pa) = .31
b. 1 ≤ P(Pa) ≤ 2
c. 0 ≤ P(Pa) ≤ 1
d. P(Pa) = .13
e. P(Pa) = .11
assume roommate B's probability assignments are correct, he would like to use the hand signal with the highest probability of winning the game. What should he use?
Rock?
To determine which method was used to assign the probabilities, we need to analyze the statement "Based on his roommate's behavior in other similar games, roommate B believes that there is a .56 probability that his roommate will choose rock and a .33 probability that his roommate will choose scissors."
If the probabilities were determined through observation and analysis of previous games, then the empirical method was used. This method relies on real-world data and experiences.
Therefore, the answer is a. empirical method.
Now, let's determine the value of P(Pa), the probability that roommate A chooses paper.
Given that the probabilities for roommate A choosing rock and scissors are provided, we can calculate the remaining probability for choosing paper.
Since there are only three options in rock-paper-scissors (rock, paper, scissors), the sum of the probabilities for all three options should be equal to 1. Therefore:
P(Pa) + P(Ra) + P(Sa) = 1
Given that P(Ra) = 0.56 and P(Sa) = 0.33, we can substitute these values into the equation:
P(Pa) + 0.56 + 0.33 = 1
P(Pa) + 0.89 = 1
Subtracting 0.89 from both sides, we find:
P(Pa) = 0.11
Therefore, the answer is e. P(Pa) = 0.11.
Now, let's determine which hand signal roommate B should use to have the highest probability of winning the game.
Since roommate A has a higher probability of choosing rock (0.56) compared to scissors (0.33), roommate B should choose paper, as paper beats rock in the rock-paper-scissors game.
Therefore, roommate B should use the hand signal of paper to maximize his chances of winning the game.