Probability question

[ryan_kidz]

Teams A nad 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?

Thnx!

 

[Gene]

A wins first 2 games .7² = 49%
B wins first 2 games .3² = 9%
The remaining 42% reqires a third game which A wins .7 so
A wins the series .49 + .7*.42 = 78.4%

 

[ryan_kidz]

woow..
thnx alot gene! u r so brilliant!

 

[soroban]

Hello, ryan_kidz!

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?

With so few possible cases, we can simply <u>list</u> the outcomes.

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.)

 

[Gene]

Soroban, I hate to pick on someone who agrees with me (even if he is dislexic) but agrees it is .784
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Gene