# Probability of winning a 5 game series by winning 3 games where probability of winning each game is given

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!

• Divide into cases, B wins the series in $3$ games, $4$ games, $5$ games. Each but the first will divide into subcases. Apr 26, 2016 at 6:11

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 $0.3*0.8*0.3 + 0.7*0.8*0.3*0.8 + .....$

• so in case $AABBB=(.7)(.2)(.3)(.8)(.8)$ is that correct?
– Mark
Apr 26, 2016 at 7:19
• Yes, that's right. Apr 26, 2016 at 8:00

Split it into disjoint events, and add up their probabilities:

• $P(AABBB)=0.7\cdot0.2\cdot0.3\cdot0.8\cdot0.8$
• $P(ABABB)=0.7\cdot0.8\cdot0.7\cdot0.8\cdot0.8$
• $P(ABBAB)=0.7\cdot0.8\cdot0.3\cdot0.2\cdot0.8$
• $P(ABBB )=0.7\cdot0.8\cdot0.3\cdot0.8$
• $P(BAABB)=0.3\cdot0.2\cdot0.7\cdot0.8\cdot0.8$
• $P(BABAB)=0.3\cdot0.2\cdot0.3\cdot0.2\cdot0.8$
• $P(BABB )=0.3\cdot0.2\cdot0.3\cdot0.8$
• $P(BBAAB)=0.3\cdot0.8\cdot0.7\cdot0.2\cdot0.8$
• $P(BBAB )=0.3\cdot0.8\cdot0.7\cdot0.8$
• $P(BBB )=0.3\cdot0.8\cdot0.3$
• thanks I made it too, and I see im correct, thanks a lot! I can check with your answer, and Im right
– Mark
Apr 26, 2016 at 7:22
• @Mark: You're welcome :) Apr 26, 2016 at 7:59