1. 1216 points | Previous Answers My NotesAsk Your Teache Teams named East and W
ID: 3361828 • Letter: 1
Question
1. 1216 points | Previous Answers My NotesAsk Your Teache Teams named East and West are playing a best-2-out-of-3 series of games. This means that the first team to win two games wins the series. (For example, in the NCAA College Baseball World Series the two teams in the finals play a best-2-out-of-3 series.) If the first two games are won by the same team, the series is over. The 3rd game is played only if the teams are 1-1 after the first two games. In each game that is played, the probability that East wins is 0.28. So the probability West wins is 0.72. What is the probability that East wins the series? 0.191296 What is the probability that the series requires only two games? 0.5968 What is the probability that the team that wins the first game wins the series? 0.7984 What is the expected value for how many games are played?Explanation / Answer
Given Probability of win a match of east =0.28
Probability of win a match of West
a)
P(series win for East) =P(first two matches won by east) +P(1st and 3 rd won by east)+P(2nd and 3rd won by east)
=(0.28*0.28) +(0.28*0.72*0.28) +(0.72*0.28*0.28)
=0.191296
P(series required only two games) =P(1st two matches won by east) +P(1st 2 matches won by west)
=0.28*0.28 +0.72*0.72
=0.5968
P(team wins the first first match will win the series) =P(1st 2mathces won by east) +(1st two matches won by west)
+P(1st and 3 rd match won by east)+P(1st and 3rd won by west)
=0.28*0.28 +0.72*0.72+0.28*0.72*0.28+0.72*0.28*0.28
=0.7984
E(no.of games played) =2*P(two match series) +3*P(3 match series)
as we know that P(2 match series) =0.5968
P(three match series) =P(1st and 3rd won by east)+P(2nd and 3 rd won by east) +P(1st and 3rd won by west)+P(2nd and 3 rd won by west )=0.28*0.72*0.28 +0.72*0.28*0.28+0.72*0.28*0.72+0.28*0.72*0.72
=0.4032
so
E(number of games played) =2* 0.5968 +3*0.4032 =2.4032
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