Question

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 an experiment. In the long run, roommate A chooses rock 24% of the time, and roommate B chooses rock 85% of the time; roommate A selects paper 12% of the time, and roommate B selects paper 14% of the time; roommate A chooses scissors 64% of the time, and roommate B chooses scissors 1% of the time. (These choices are made randomly and independently of each other.)

The probabilities were assigned using the:_________

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

Answer 1

The probabilities are assigned using the empirical probability (experimental).

`$P(A) =P(\text{A rock and B scissor +A scissor and B paper+A paper and B rock})$`

`$=(0.24*0.01 + 0.64 * 0.14 + 0.12 * 0.85) $`

= 0.194

`$P(C) =P(\text{both rock + both paper + both scissor})$`

`$=(0.24*0.85+ 0.12 * 0.14 + 0.64 * 0.01) $`

= 0.227

`$P(B)=1 - P(A)-P(C)$`

= 1 - 0.194 - 0.227

= 0.579

≈ 0.58

∴
`$A^C = $` event of B or event of C

So the probability of
`$A^C $` is 0.81