union of two independent probabilistic event

If the events $A$ and $B$ are independent, then $P(A \cap B) = P(A) P(B)$ and not necessarily $0$.

You are confusing independent with mutually exclusive.

For instance, you toss two coins. What is the probability that both show heads? It is $\frac{1}{2} \times \frac{1}{2}$ isn't it? Note that the coin tosses are independent of each other.

Now you toss only one coin, what is the probability that it shows both heads and tails?

If $A$ and $B$ are 2 independent events then :

\begin{align*} P(A \cup B) &= P(A) + P(B) - P(A)\cdot P(B) \quad \Bigl[\because \tiny P(A \cap B) = P(A) \cdot P(B) \ \text{for independent events} \Bigr] \\ &= \frac{4}{10} + \frac{7}{10} - \frac{28}{100} \\ &= \frac{110-28}{100} = \frac{82}{100} =0.82 \end{align*}

Please refer this link:

• http://www.proofwiki.org/wiki/Definition:Independent_Events

No, the last one is not $0$. If $A,B$ are independent, then $P(A \cap B) = P(A) P(B) = 0.4 \cdot 0.7 = 0.28$