1. independent

P(A or B) = .60 and P(A) =.20

How do I find P(B) given that A and B are independent?

Since I see that we have an "or" and not an "and," I would add up the probabilities so P(B)=.40? This is also the probability if A and B were mutually exclusive too?

2. Hi,

This is my first post so you may want to check it with someone who knows something about this

Anyway I googled the problem and it came up with this

P(A or B) = .60 and P(A) =.20 ... How do I find P(B) given that A and B are independent? - Yahoo! Answers

The trouble is that it gives more than one answer .5 and .4 are both suggested. Personally I find .5 more convincing. Anyway below is the best argument I found on the link.

There's a 40% chance of not A or B
There's an 80% chance of not A
Thus there's a 40% chance of B but not A

Now, given not A, there's a 50% chance of B (40%/80%)
Since A and B are independent, there must be a 50% chance of B regardless of the condition of A.

3. Hello, sfspitfire23!

$P(A \cup B) \,=\, 0.60,\;\;P(A) \,=\,0.20$

Given that $A$ and $B$ are independent, find $P(B).$

We have this formula: . $P(A \cup B) \:=\: P(A) + P(B) - P(A \cap B)$

Since $A$ and $B$ are independent: . $P(A \cap B) \:=\:P(A)\!\cdot\!P(B)$

The formula becomes: . $P(A \cup B) \;=\;P(A) + P(B) - P(A)\!\cdot\!P(B)$

Substitute: . $P(A \cup B) = 0.6,\;\;P(A) = 0.2$

. . $0.6 \;=\;0.2 + P(B) - 0.2\!\cdot\!P(B)$

. . $0.4 \;=\;0.8\!\cdot\!P(B)$

Therefore: . $P(B) \;=\;\frac{0.4}{0.8} \;=\;0.5$