Given, P(E1)=0.25,P(E2)=0.50
and P(E1andE2)=P(E1∩E2)=0.14.
∴P(
¯
E1
)=1-P(E1)=(1-0.25)=0.75
and P(
¯
E2
)=1-P(E2)=(1-0.50)=0.50.
∴P(neither E1 nor E2)
=P( not E1 and not E2)=P(
¯
E1
∩
¯
E2
)
=P(
¯
E1∪E2
)={1-P(E1∪E2)}
=1-{P(E1)+P(E2)-P(E1∩E2)}
=1-(0.25+0.50-0.14)=(1-0.61)=0.39.
Hence, the required probability is 0.39.
