Conditional probabilties with multiple events.

Question

Define the following events:

$A:$ Firm owns a word processor

$B1:$ Firm's sales are less than $500,000\$$ a year.

$B2:$ Firm's sales are $500,000 \$$ to under $10$ million a year.

$B3:$ Firm's sales are $10\$$ million or more a year.

Assume that:

$P(A)= 0.16$ $P(B1)= 0.3$ $P(B2)= 0.6$ $P(A|B2)= 0.1$ $P(A|B3)= 0.9$

$a)$ Find $P(A \cap B2)$ Find $P(A \cup B3)$ Find $P(B3|A)$

$b)$ Are owning a word processor and the firm's level of sales independent? If so, explain how you reached this conclusion. If not, describe the relationship between them.

I have found the $P(A \cap B2)= 0.06$. I cannot find the rest.

Answer 1

$P(A \cap B_2)=P(B_2) \times P(A|B_2)=0.6 \times 0.1=0.06$

In Addition $\sum P(B_i)=1$ since they partition all possibilities. So $P(B_3)=0.1$

$P(A|B_3)=P(A \cap B_3)/P(B_3)$ So $P(A \cap B_3)=0.9 \times 0.1$

$P(A \cup B_3)=0.16+0.1-0.09=0.17$

For third part you need Bayes formula

$P(B_3|A) \times P(A)= P(A|B_3) \times P(B_3)$

$P(B_3|A) \times 0.16= 0.9 \times 0.1$

So $P(B_3|A) =0.5626$

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For second question the answer is NO. Because $P(A)

So knowing $B_3$ increases the probability of $A$ hence they are dependent.