We have
P(E1)=0.5,P(E2)=0.3andP(E1∩E2)=P(E1andE2)=0.1
.
∴P(
¯
E1
)={1-P(E1)}=(1-0.5)=0.5,
and P(
¯
E2
)={1-P(E2)}=(1-0.3)=0.7.
Thus, we have
(i) P(E1 or E2)=P(E1∪E2)
=P(E1)+P(E2)-P(E1∩E2)
=(0.5+0.3-0.1)=0.7.
(ii) P(E1 but not E2)=P(E1∩
¯
E2
)
=P(E1)-P(E1∩E2)
=(0.5-0.1)=0.4.
(iii) P(E2 but not E1)=P(E2∩
¯
E1
)
=P(E2)-P(E2∩E1)
=P(E2)-P(E1∩E2)=(0.3-0.1)=0.2.
(iv) P(neither E1 nor E2)=P( not E1 and not E2)
=P(
¯
E1
and
¯
E2
)=P(
¯
E1
∩
¯
E2
)
=P(
¯
E1∪E2
)=1-P(E1∪E2)
=1-P(E1∪E2)
=(1-0.7)=0.3
[using (i) ].