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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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    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)$2011-07-16

1 Answers 1

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The formula is the same for both. The probability of their union is the sum of their (individual) probabilities, minus the probability of their intersection.

In a) their intersection is 0.


For the second problem, we have the exact same principle at work: add their individual probabilities (0.92 + 0.92), and subtract the probability that she inherited the gene from both parents (i.e. 0.92^2).

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    Not intuitive. But mindlessly plugging in numbers in the wrong formula during the exam could happen. I understand it now, there are three outcomes where she is cblind and one where she isn't. Thanks for the help.2011-07-16