(2 points) A professional Rock, Paper, Scissors player can win a single game (tr
ID: 3065128 • Letter: #
Question
(2 points) A professional Rock, Paper, Scissors player can win a single game (trial) with probability 0.88. The player is up against an opponent. What is the probability that the player will win 2 games before the opponent wins 2 games? Hint: You want to add several values of a negative binomial distribution u(m, k,p) with k being the number of games needed to win and p being the probability of winning What values of m (the number of games played) do you want? Can the player get 2 wins in 1 game? Can the player get 2 wins over 2 games without the opponent getting 2 wins? How about 3 games? 4? What is the probability that the player will win 1 game before the opponent wins 2 games?Explanation / Answer
1) probability of winning 2 game before oppponent wins twice =P(player win twice +opponent wins and then player wins and then player wins+player then opponent and then palyer) =0.88*0.88+0.12*0.88*0.88+0.88*0.12*0.88
=0.960256
2)probability=P(player wins+opponent wins then player wins) =0.88+0.12*0.88 =0.9856
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