Two teams play a series of baseball games, team A and team B. The team that wins 3 of 5 games wins the series. The first game takes place in the stadium of the team A, the second in the stage of team B, and the third stage in the stadium of team A, and if reaching a fourth and fifth games, they were held in the stadium of the team B.
It is known that when playing at their stadium the team A, has a chance to beat the team B equals to 0.7, while when played at the stadium of the team B, the probability that team A will win the team B is equal to 0.2. Assuming the match results are independent of each other, calculate the probability that B wins the series.
Hi to everybody, I dont how to proced in this problem,can someone help? thanks!
$\endgroup$ 12 Answers
$\begingroup$You will have to sum up a number of mutually exclusive probabilities:
The last winning match has to be $B$, and $B$ must win any $2$ of the preceding ones,
hence $\binom22 + \binom32 + \binom42$ cases
$BBB,\;ABBB,\; BABB,\; BBAB,\; AABBB,\; ABABB,\; ......$
with probabilities $0.3*0.8*0.3 + 0.7*0.8*0.3*0.8 + .....$
$\endgroup$ 2 $\begingroup$Split it into disjoint events, and add up their probabilities:
- $P(AABBB)=0.7\cdot0.2\cdot0.3\cdot0.8\cdot0.8$
- $P(ABABB)=0.7\cdot0.8\cdot0.7\cdot0.8\cdot0.8$
- $P(ABBAB)=0.7\cdot0.8\cdot0.3\cdot0.2\cdot0.8$
- $P(ABBB )=0.7\cdot0.8\cdot0.3\cdot0.8 $
- $P(BAABB)=0.3\cdot0.2\cdot0.7\cdot0.8\cdot0.8$
- $P(BABAB)=0.3\cdot0.2\cdot0.3\cdot0.2\cdot0.8$
- $P(BABB )=0.3\cdot0.2\cdot0.3\cdot0.8 $
- $P(BBAAB)=0.3\cdot0.8\cdot0.7\cdot0.2\cdot0.8$
- $P(BBAB )=0.3\cdot0.8\cdot0.7\cdot0.8 $
- $P(BBB )=0.3\cdot0.8\cdot0.3 $
