Sue and Mary agree to play the best of three games of badminton. They stop if either of them have won two games. Over a long period of time, Sue wins 4 games out of every 5 games they play. Find the probability that Sue will win the best of three.
I have that the chance of winning best of three is 0.50 but not sure what next.?
Prob(m) = 1/5
prob(S) = 4/5
Possible outcomes in best of three"
MM --- (1/5)(1/5) = 1/25
MSM -- ((1/5)(4/5)(1/5) = 4/125
SMM -- ((4/5)(1/5)(1/5) = 4/125
SS ---- (4/5)(4/5) = 16/25 ----> Sue wins
SMS -- (4/5)(1/5)(4/5) = 16/125 ----> Sue wins
MSS -- ((1/5)(4/5)(4/5) = 16/125---> Sue wins
prob(sue wins) = 16/25+16/125+16/125 = 112/125 or .896
add them up to get 1 as a check
you said "I have that the chance of winning best of three is 0.50"
How so ??
To find the probability that Sue will win the best of three games, you can break it down into three possible scenarios: Sue wins in two games, Sue wins in three games, or Sue loses.
1. Scenario 1: Sue wins in two games.
In this case, Sue wins the first two games and the match is over. The probability of Sue winning any particular game is 4/5, so the probability of her winning both games is (4/5) * (4/5) = 16/25.
2. Scenario 2: Sue wins in three games.
In this case, Sue wins the first game, Mary wins the second game, and Sue wins the third game. The probability of Sue winning a game is 4/5, and the probability of Mary winning a game is 1/5. So, the probability of this scenario is (4/5) * (1/5) * (4/5) = 16/125.
3. Scenario 3: Sue loses.
In this case, Mary wins the first two games and the match is over. The probability of Mary winning any particular game is 1/5, so the probability of her winning both games is (1/5) * (1/5) = 1/25.
To calculate the overall probability that Sue will win the best of three, you need to sum up the probabilities of the three scenarios: (16/25) + (16/125) + (1/25) = 91/125.
Therefore, the probability that Sue will win the best of three games is 91/125, which is approximately 0.728 or 72.8%.