1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Do you know what it means for $A$ and $B$ to be mutually exclusive? What about for $C$ and $D$ to be independent? $\endgroup$ – JavaMan Jun 10 '13 at 12:06
3
$\begingroup$

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 $P(D)$.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
  • 2
    $\begingroup$ "so the logic for the previous answer given its not correct" Sorry? $\endgroup$ – Did Feb 27 '14 at 23:35
-2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ $P(C\cap D)$ is not necessarily zero for independent events. I think you're confusing independent with mutually exclusive. $\endgroup$ – Stefan Hansen Aug 13 '13 at 11:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.