Here it is given that he probability that event A occurs is \(P(A)=0.3\), the probability that event B does not occur is 0.4 that is \(\displaystyle{P}{\left({B}^{{c}}\right)}={0.4}\) therefore \(\displaystyle{P}{\left({B}\right)}={1}-{P}{\left({B}^{{c}}\right)}={1}-{0.4}={0.6}\)

For two event A and B, we know that \(\displaystyle{P}{\left({A}∪{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}⋂{B}\right)}\).

Now if \(P(A∪B)=0.7\). therefore

\(\displaystyle{P}{\left({A}⋂{B}\right)}={P}{\left({A}\right)}+{P}{\left({B}\right)}-{P}{\left({A}∪{B}\right)}=0.3+0.6-0.7 =0.9-0.7 =0.2\)

Therefore the probability that events A and B occur simultaneously is 0.02