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Probability question
- Thread starter ryan_kidz
- Start date
Hello, ryan_kidz!
There are three ways that A wins the series:
. . A wins the first two games: AA
. . A wins the first and third games: ABA
. . A wins the second and third games: BAA
\(\displaystyle P(AA)\ =\ (0.7)^2\ =\ 0.49\)
\(\displaystyle P(ABA)\ =\ (0.7)(0.3)(0.7)\ =\ 0.147\)
\(\displaystyle P(BAA)\ =\ (0.3)(0.7)(0.7)\ = \ 0.147\)
\(\displaystyle P(A\text{ wins series}) \;= \;0.49\,+\,0.147\,+\,0.147 \;= \;0.748\)
. . . (But I prefer Gene's explanation.)
With so few possible cases, we can simply <u>list</u> the outcomes.Teams A and B play a series of games; whoever wins two games first wins the series.
If team A has a 70% chance of winning any single game, what is the probability that Team A wins the series?
There are three ways that A wins the series:
. . A wins the first two games: AA
. . A wins the first and third games: ABA
. . A wins the second and third games: BAA
\(\displaystyle P(AA)\ =\ (0.7)^2\ =\ 0.49\)
\(\displaystyle P(ABA)\ =\ (0.7)(0.3)(0.7)\ =\ 0.147\)
\(\displaystyle P(BAA)\ =\ (0.3)(0.7)(0.7)\ = \ 0.147\)
\(\displaystyle P(A\text{ wins series}) \;= \;0.49\,+\,0.147\,+\,0.147 \;= \;0.748\)
. . . (But I prefer Gene's explanation.)