(2 points) A professional Rock, Paper, Scissors player can win a single game (trial) with probability 0.86. The player is up against an opponent. What is the probability that the player will win 2 games before the opponent wins 2 games

The probability that the player wins 2 games before the opponent wins two games is 0.9466.

Step-by-step explanation:

The probability that a player (P) wins a single professional Rock, Paper, Scissors game is p = 0.86.

Then the probability that the opponent (O) wins a game is,

q = 1 - p

  = 1 - 0.86

  = 0.14

The sample space such that the player wins two games before the opponents wins two games is:

S = {PP, POP, OPP}

Compute the probability of each event as follows:

                                                            × P (Players wins 1st game)

                                                                × P (Opponent wins 2nd game)

                                                                        × P (Player wins 3rd game)

                                                                × P (Player wins 2nd game)

                                                                        × P (Player wins 3rd game)

Compute the probability that the player wins 2 games before the opponent wins two games as follows:

P (Player wins 2 games before the opponent) = P (PP) + P (POP) + P (OPP)

Thus, the probability that the player wins 2 games before the opponent wins two games is 0.9466.

not solvable

step-by-step explanation:

we need more information, or at least an

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!