Question

In a netball series, teams A and B play each other until one team has won 4 games. If team A has a 2/3 probability of winning against B in a single game, what is the probability that the series will end after exactly 7 games have been played?

Answer 1

For it to last exactly 7 games:

P(A win in 7 games) + P(B win in 7 games)

P(A win in 7 games) = P(AAABBBA)+P(BBBAAAA)+P(ABABABA)+P(AABBABA) + P(BBAAABA) + P(BBAABAA)

There are some other combinations of 4 A and 3 B, Try to think of some more but keep in mind that the last should be A win

Removing that one A,

You have 3 A and 3 B to arrange = 6 ! / (3! x 2! )

Therefore, P (A wins in 7 matches) = (2/3)^4 x (1/3)^3 x 6 ! / (3! x 2! )

The same applies for B wins,

Therefore,

P(Series last exactly 7 Matches) = (2/3)^4 x (1/3)^3 x [ 6 ! / (3! x 2! ) ] x 2 = 0.439 (to 3 s.f)

Answer 2

@Kropot72 If you say 7C4, it includes the combination AAAABBB but in this case, The series ends at 4 matches since A already won the 4 required matches, so i don't think you can use that :)

Answer 3

yea

Answer 4

@sk8teroy to make sample space do this
S = {A,B}x{A,B}x{A,B}x{A,B}x{A,B}x{A,B}x{A,B}
S = {AA,AB,BA,BB} x {AA,AB,BA,BB} x {AA,AB,BA,BB} x {A,B}
S = {AAAA,AAAB,AABA,AABB,....,BBBA,BBBB} x {AAA,AAB,ABA,ABB,BAA,BAB,BBB}
S = {AAAAAAA,AAAAAAB,.....,BBBBBAB,BBBBBBB}
make a complete sample space then ignore all the combinations in which each team win 4 matches together...then calculate the probability of remaining events

Answer 5

Calculate the probabilities of team A and team B winning 3 out of the first 6 games of the 7. Then multiply those probabilities by the probability of team A winning the seventh game and team B winning the seventh game.
\[P(team\ A\ wins\ 3\ out\ of\ 6)=\left(\begin{matrix}6 \\ 3\end{matrix}\right)\times(\frac{2}{3})^{3}\times(\frac{1}{3})^{3}=0.21948\]
\[P(team\ A\ wins\ series)=0.21948\times\frac{2}{3}=0.14632\]
\[P(team\ B\ wins\ series)=0.21948\times\frac{1}{3}=0.07316\]
The probability that the series will end after exactly 7 games have been played is:
0.14632 + 0.07316 = 0.21948

Answer 6

Note that the probability that team A wins 3 out of 6 is the same as the probability that team B wins 3 out of 6.