View Full Version : Teams A and B take turns playing for championship....

bbash30

04-08-2008, 06:20 PM

Suppose that two teams, A and B, are playing each other for the championship and that team A has a 3/5 probability of winning any game between the two teams.

A] If the championship is best three of five, what is the probability that team A wins?

B] If the championship is best four of seven, what is the probability that team A wins?

would a tree work?

Suppose that two teams, A and B, are playing each other for the championship and that team A has a 3/5 probability of winning any game between the two teams.

A] If the championship is best three of five, what is the probability that team A wins?

B] If the championship is best four of seven, what is the probability that team A wins?

For part A

\sum\limits_{k = 3}^5 {{\binom {5}{k}}\left( {\frac{3}{5}} \right)^k \left( {\frac{2}{5}} \right)^{5 - k} }

Make a simple change for part B.

bbash30

04-08-2008, 08:12 PM

( 5!/3!(2!) ) (.216) (.16) = .3456 (However, I am not positive on the "sum" sign, what step does that involve?

Does this equation work or should I write out all possible outcomes?

In the following graphic you can see the answer.

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