# How do I find the probability of these independent events

The following is taken from the ETS math review for the GRE:

Let A, B, C, and D be events for which P(A or B)=0.6, P(A)=0.2, P(C or D)=0.6, and P(C)=0.5 The events A and B are mutually exclusive, and the events C and D are independent. Find P(D).

I'd appreciate it if someone could show me how to solve this. The answer is given as 0.2, but I don't know how it's arrived at. Thanks!

• Do you know what it means for $A$ and $B$ to be mutually exclusive? What about for $C$ and $D$ to be independent? – JavaMan Jun 10 '13 at 12:06

Using the independence of $C$ and $D$ we have that $$P(C)P(D)=P(C\cap D)=P(C)+P(D)-P(C\cup D).$$ Use this to find $P(D)$. You don't need to involve $A$ and $B$ to find $P(D)$.

This is a question from the GRE practice book and the answer is not $0.1$, so the logic for the previous answer given ([edit LL:] by user90188) its not correct. I'm almost sure the answer is:

b) $P(C\text{ or }D) = P(C)+P(D)-P(C\text{ and }D)$ since independent events may be NOT mutually exclusive events. \begin{align} P(C\text{ or }D) &= P(C)+P(D)-P(C)*P(D) \\ 0.6 &= 0.5+P(D)-0.5*P(D) \\ 0.6-0.5&=P(D)-0.5*P(D) \\ 0.1&=P(D)-0.5*P(D) \\ 0.1&=P(D)(1-0.5) \\ 0.1/(1-0.5)&=P(D) \\ 0.1/0.5&=P(D) \\ 0.2&=P(D) \\ \end{align}

• "so the logic for the previous answer given its not correct" Sorry? – Did Feb 27 '14 at 23:35

P(c∩d)= 0 as they are independent so

p (c∪d)= p(c)+p(d)-P(c∩d)

so p(d)= 0.6-0.5

p(d)=0.1

• $P(C\cap D)$ is not necessarily zero for independent events. I think you're confusing independent with mutually exclusive. – Stefan Hansen Aug 13 '13 at 11:16