SOLUTION: Suppose that two teams, A and B, play each other five times over a season (not best of five series, but they play all five games). If Team A has a .6 probability of winning any giv
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-> SOLUTION: Suppose that two teams, A and B, play each other five times over a season (not best of five series, but they play all five games). If Team A has a .6 probability of winning any giv
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Question 1195025: Suppose that two teams, A and B, play each other five times over a season (not best of five series, but they play all five games). If Team A has a .6 probability of winning any given game complete the probability distribution below (2 points):
XA P(XA)
0 .01
1 .08
2
3
4
5
Total 1.0
What is the probability that Team A sweeps the season series (wins all five matchups)?
What is the probability that Team A wins at least three games?
What is the probability that Team A gets swept (does not win a game)? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! winning none is 0.4^5=0.01 given
winning one is 0.4^4*0.6*5 ways=0.08 rounded
winning two is 5C2*0.4^3*0.6^2=0.23
winning three is 5C3 *0.4^2*0.6^3=0.346
winning four is 5C4*0.4*0.6^4=0.26
winning all is 0.6^5=0.08
adds up to 1.006 because of rounding.
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wins at least 3 is 0.686 or 0.69
gets swept is 0.01, given above
