a)
If i is to be 4 and we wish to find the probability of playing7 games overall, then there must have been 6 games in which eachteam won 3
P(3 in 6) = 6C3 (p)3*(1-p)3 =20 (p3) (1 - 3p + 3p2 - p3)
Now we need to use this and consider each of the cases whereteam A wins the 7th game and where Team B wins the 7th game.
P(7 games) =P(6 games and A wins the 7th) +P(6 games and Bwins the 7th)
= p*(20p3)(1-3p+3p2-p3)+ (1-p)*(20p3)(1-3p+3p2-p3)
b)
If i = 3
Number of chances that both teams are win three games
AAA AABA AABBA
BBB ABAA ABBAA
BAAA BBAAA
BBAB BBAAB
BABB BBABB
ABBB BABAB
AABBB
ABABA
i.e if any of the team must win three games, then the game stops.
The probability that A wins is p
The probability that B wins is = 1-p = q
Therefore i = 3 the expected number of games played is
E(x) = 3[p*p*p + q*q*q + 4(p*p*q*p + p*q*p*p + q*p*p*p + q*q*p*q + q*p*q*q + p*q*q*q) + 5(p*p*q*q*p +p*q*q*p*p + q*q*p*p*p + q*q*p*p*q + q*p*p*q*q + p*p*q*q*q + q*p*q*p*q + p*q*p*q*p)]
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