# how to calculate discrete event probability?

A bookstore sells children's books that belong to two publishing companies A and B and were published between years 2000 and 2004. The probabilities of a book being published by companies A and B are 0.6 and 0.4. The probability that company A published a book in years 2000, 2001, 2002, 2003, 2004 are 0.1, 0.2, 0.3, 0.3 and 0.1 respectively. The probability that company B published a book in years 2000, 2001, 2002, 2003, 2004 are 0.3, 0.2, 0.1, 0.2 and 0.2 respectively.

a) Find the probability that a book was published after 2002 b) Find the probability that a book published after 2002 was published by company A.

I think that for solution a)

probability that a book was published after 2002 is

\begin{align} P(A \cup B) &= P(A) + P(B) - P(A)P(B) \end{align}

which is equal to

\begin{align} P(A \cup B) &= \frac{4}{10}+\frac{4}{10}-\frac{4}{10}\frac{4}{10} = 0.64 \end{align}

for solution b) i don't understand the question exactly but

If I assume that it asks only the probability that a book published after 2002 was published by company A

\begin{align} P(A \cup \overline{B}) &= P(A) + P(\overline{B}) - P(A)P(\overline{B}) \end{align}

which is equal to

\begin{align} P(A \cup B) &= \frac{4}{10}+\frac{6}{10}-\frac{4}{10}\frac{6}{10} = 0.76 \end{align}

Otherwise if I assume that only the probability that a book published after 2002 was published by company A and the probability that a book published after 2002 was published by company A and B

So the solution is

$$P(A) = 4/10 = 0.4$$

Is solution of a is correct and for b which solution is correct?

$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Company}& \text{2000}& \text{2001}& \text{2002}& \text{2003}& \text{2004} \\ \hline \text{A} & .06& .12& .18& .18& .06 \\ \hline \text{B} & .12& .08& .04& .08& .08 \\ \hline \end{array}$$