(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.

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step-by-step explanation:

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