The probability that Team A will score 2 points is p1= C32× (23) 2× (1-23) = 49.
The scoring probability of team B 1 is P2 = 23× (1-23 )× (1-kloc-0/2)+(1-23 )× 23× (1-kloc).
Then p (a) = 49× 518 =1081;
The probability that Team A gets 3 points P3=C33×(23)3=827.
The probability of team B's zero P4 = (1-23) × (1-23 )× (1-kloc-0/2) =1/8.
Then p (b) = 827×118 = 4243;
(2) From the meaning of the question, the acceptable value of ξ is 0, 1, 2, 3,
ξ=0, that is, Team A scored 0, and all three people got it wrong, p (ξ = 0) = C03× (1? 23)3= 127,
ξ= 1, that is, Team A scored 1, among which only 1 was correct, and P (ξ =1) = C13× 23× (1? 23)2=29,
ξ=2, that is, Team A scored 2 points, and only 2 out of 3 people answered correctly, P (ξ = 2) = C23× (23) 2× (1? 23)=49,
ξ=3, that is, Team A scored 3 points, and all three people got it right, P (ξ = 3) = C33× (23) 3 = 827.
So the distribution list of ξ is
The mathematical expectation of ξ is e ξ = 0×127+1× 29+2× 49+3× 827 = 2.