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

Solution :

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

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

        $=(0.24\times0.01 + 0.64 \times 0.14 + 0.12 \times 0.85)$

        = 0.194

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

        $=(0.24\times0.85+ 0.12 \times 0.14 + 0.64 \times 0.01)$

        = 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