Let P(S1)P(S2)P(S3)===abc they are independently solvingU=S1∪S2∪S3⇒UC=(S1∩S2∩S3)⇒P(U)=1−P(Uc)=1−[(1−a)(1−b)(1−c)]=21⇒(1−a)(1−b)(1−c)=21P(V)=P(S2∩S3S1)=P(S2∩S3)P(S1∩S2∩S3)=P(s1)=a=101P(W)=P(S2∩S3)=b(1−c)=121(1−101)(1−b)(1−c)=21⇒(1−b)(1−c)=951−bb=121×59=203⇒b=233233(1−c)=121⇒1−c=3623⇒c=3613P(T)=P(S3)=c=3613