# Simple probability question; why the calculation is as it is

Well the task is as follows.
About two events A and B, we know that $P(A)=\frac{1}{6}$ and $P(B)=\frac{2}{7}$. Find $P(A \cup B)$
a) If A and B is disjunktive
b) If A and B is independent

My solutions
a) $\frac{1}{6}+\frac{2}{7}$
b) $\frac{1}{6}+\frac{2}{7} - (\frac{1}{6} \times \frac{2}{7})$
This gives the correct answer but I'm not 110% sure why.

And, the same problem with another task.

For a girl to be colorblind both her parents needs to carry the gene for it. She inherits the gene from both parents independently. The probability is 8% that the girl gets that gene from the mother and 8% for she to get it from the father.
What is the probability that one girl is colorblind?
What is the probability that one girl is not colorblind?

The first one is just: $0.08^2$. That's fine.
The second one is $0.92+0.92 - 0.92^2$

I don't understand why the second isn't just $0.92^2$ like the first one.

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(+1) because I want to see more of such users and questions on this site. –  The Chaz 2.0 Jul 15 '11 at 20:48
Thanks for the upvote. I need to understand this 110%, I need an A in this class. –  Algific Jul 15 '11 at 21:15
Just remember that understanding, like probability, can never exceed 100%!  ;) –  The Chaz 2.0 Jul 15 '11 at 21:41
I wish I could give more than one upvote for that comment. –  mathmath8128 Jul 15 '11 at 21:45
For b), use that if A,B are independent, then P(A|B)=P(A)=$\frac {P(A\cap B)}{P(B)}$ , so that $P(A)P(B)=P(A\cap B)$ , then $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ –  gary Jul 16 '11 at 0:28