# Statistics/Probability

2. Two roommates, roommate X and roommate Y, are expecting company and are arguing over who should have to wash the dishes before the company arrives. Roommate X suggests a game of rock-paper-scissors to settle the dispute.
Consider the game of rock-paper-scissors to be a random experiment. In past games, roommate X has chosen rock 36% of the time, and roommate Y has chosen rock 22% of the time; roommate X has selected paper 32% of the time, and roommate Y has selected paper 25% of the time; roommate X has chosen scissors 32% of the time, and roommate Y has chosen scissors 53% of the time. (These choices are made randomly and are independently of each other.) Use these percentages to assign probabilities to the roommates' hand signal choices.
The probabilities were assigned using the
a. empirical method
b. theoretical method
c. subjective method
A?
Define event Y as the event that roommate Y wins the game and thus does not have to wash the dishes.
What is P'(Y), the probability of event Y?
a. .32
b. .33
c. .43
d. .22
I don't even know how I'm supposed to calculate that
Let event W be the event that the game ends in a tie.
What is P'(W), the probability of event W?
a. .32
b. .44
c. .55
d. .33
Again, idk!

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1. Define event X as the event that roommate X wins the game and thus does not have to wash the dishes. What is the probability that roommate X wins the game?
a. .34
b. .37
c. .67
d. .43
What is the complement of event Y?
a. event Y or event W
b. event X or event W
c. event Y
d. event W
I think it is B
The probability of the complement of event Y is
a. .34
b. .61
c. .13
d. .67

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2. A is correct.

I assume you know that rock wins over scissors, paper wins over rock, and scissors wins over paper.

For the other parts, make a tree diagram complete with corresponding probabilities and sum all the appropriate outcomes to get the probability.
Example:

R=rock, P=paper, S=scissors
P(AB): probability that X chooses A, Y chooses B

The probability tree for the complete game is as follows:
cases
1.P(RR)=0.36*0.22=0.0792
2.P(RP)=0.36*0.25=0.0900
3.P(RS)=0.36*0.53=0.1908
4.P(PR)=0.32*0.22=0.0704
5.P(PP)=0.32*0.25=0.0800
6.P(PS)=0.32*0.53=0.1696
7.P(SR)=0.32*0.22=0.0704
8.P(SP)=0.32*0.25=0.0800
9.P(SS)=0.32*0.53=0.1696
Total probabilities=1.0

For Y to succeed, add the probabilities P(RP)+P(PS)+P(SR)=0.0900+0.1696+0.0704
=0.3300

You can solve the remaining cases similarly.

B is correct, because X,Y and W make the sample space and the probabilities should add to 1.

The probability of Y' (complememnt of Y) , i.e. P(Y') can be obtained by subtracting P(Y) from 1.
P(Y')=1-P(Y).

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3. Thanks a lot that made it simpler!
So for the probability of W, would it be .32 or .33?

Could you please look at my other posts as well and see if you can make them a little more clear for me please!

Again, thanks a lot!

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4. And please disregard the probability of W question, It's .33 I'm pretty sure

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5. The sum
P(RR)+P(PP)+P(SS)=0.3288 which rounds to 0.33.

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