Respuesta :
Answer:
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:
- Probability of the event PP = P (Players wins 1st game)
× P (Players wins 1st game)
[tex]=0.86\times0.86\\=0.7396[/tex]
- Probability of the event POP = P (Players wins 1st game)
× P (Opponent wins 2nd game)
× P (Player wins 3rd game)
[tex]=0.86\times0.14\times0.86\\=0.103544\\\approx0.1035[/tex]
- Probability of the event OPP = P (Opponent wins 1st game)
× P (Player wins 2nd game)
× P (Player wins 3rd game)
[tex]=0.14\times0.86\times0.86\\=0.103544\\\approx0.1035[/tex]
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)
[tex]=0.7396+0.1035+0.1035\\=0.9466[/tex]
Thus, the probability that the player wins 2 games before the opponent wins two games is 0.9466.