Question

随机事件A、B、C两两相互独立且ABC=空集,且P(A)=P(B)=P(C)

Answer 1

已知:P(AUBUC)=P(A)+P(B)+P(C)- P(AB)- P(AC)- P(BC) +P(ABC).

对于本题,P(ABC)=0,P(AB)=P(A)*P(B) ,P(AC) = P(A)P(C),P(BC)= P(B)*P(C)

有:P(AUBUC)=P(A)+P(B)+P(C)- P(A)P(B)- P(A)P(C)- P(B)P(C)

再由:P(A)=P(B)=P(C).及P(AUBUC)=9/16,得

3P(C)- 3[P(C)]^2 = 9/16.记P(C)=x,即有方程式:3x^2-3x+9/16 =0

或:x^2 -x +3/16=0.解得:

x1=[1-1/2]/2=1/4,x2= [1+1/2]/2=3/4.由于假设P(C)