Nishant A. answered 10/07/23
MATHS IS EASY GIVE IT A FAIR CHANCE
A,B,C are independent events implies
P(A∩B)=P(A).P(B)
P(B∩C)=P(B).P(C)
P(C∩A)=P(C).P(A)
P(A∩B∩C)=P(A).P(B).P(C)
now we have top prove P(A∩(B∪C))=P(A).P(B∪C)
solving R.H.S.
P(A).P(B∪C)=P(A)[(P(B)+P(C)-P(B∩C)]
P(A)[P(B)+P(C)-P(B).P(C)]
P(A).P(B)+P(A).P(C)-P(A).P(B).P(C)
P(A∩B)+P(A∩C)-P(A∩B∩C)
P(A∩(B∪C))
hence proved