# A and B are mutually exclusive, C and D are independent

Another GRE study question

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.

Part (a) asks find P(B), which is

$P(A \cup B) = P(A) + P(B)$ $0.6 = 0.2 + P(B)$ $P(B) = 0.4$

But part (b) asks find P(D), and when I try, my answer is $0.1$

$P(D) = P(C\cup D) - P(C) = 0.6 - 0.5 = 0.1$

This is incorrect. According to the study guide, answer is $0.2$

• Subtract $P(C\cap D^c)$ rather than $P(C)$. Jan 1, 2016 at 19:43

$$P(C\cup D) = P(C)+P(D)-P(C\cap D)$$

$$0.6 = 0.5 +P(D) - P(C).P(D) = 0.5 +P(D) - 0.5P(D)$$

$$.5P(D) = .1$$

$$P(D) = .2$$

The catch is Cand D are independent, then $P(C\cap D$ = P(C).P(D)

• When calculating union of C and D, i.e. $P(C∪D)$, why are you subtracting the intersection of C and D, i.e. $P(C∩D)$ Jan 1, 2016 at 20:46
• The property of Mutually Exclusive Events means that there is nothing common in them. For those that are not mutually exhaustive, you have something in the common which is $\cap$. Axiom says that C or D must be equal to All that contain in C + All that contain in D - all that contain in C and D ( to remove double counting) and hence the probability. The other condition of indenpendence is always applied on $\cap$ in a way $P(C and D) = P(C\cap D) = P(C).P(D)$. Jan 1, 2016 at 21:02
• You are welcome!! Jan 1, 2016 at 21:29

a) OR$=\cup$, AND$=\cap$ Since $A$ and $B$ are disjoint (mutually exclusive), then $$\{A\cap B\} = \{AB\} = \varnothing.$$ Thus $$P(AB) = P(\varnothing) = 0.$$ Recall the inclusion-exclusion rule $$P(A\cup B) = P(A)+P(B) - P(AB).$$ This implies $$P(B) = P(A\cup B)-P(A)+P(AB) = .6-.2+0 = .4$$

b) Since, $C$ and $D$ are independent, $$P(CD) = P(C)P(D).$$ Again, by inclusion-exclusion, \begin{align*} P(D) &= P(C\cup D) -P(C)+P(CD)\\ &= P(C\cup D)-P(C)+P(C)P(D), \end{align*} and combining like terms yields $$P(D)-P(C)P(D) = P(D)[1-P(C)] = P(C\cup D)-P(C).$$ Solving for $P(D)$ gives $$P(D) = \frac{P(C\cup D)-P(C)}{1-P(C)} = \frac{.6-.5}{1-.5} = .2.$$

You can't say $P(D)=P(C$ or $D)-P(C)$ because $C$ and $D$ are not mutually exclusive. Independent implies that they are not mutually exclusive.

• How is mutually exclusive different from independent. Please explain. Jan 1, 2016 at 19:44
• Two events are mutually exclusive if $A\cap B=\emptyset$. So for example suppose you flip a coin, two outcomes $H$ and $T$. Let $A$ be the event $\{H\}$ and let $B$ be the event $\{T\}$. Then they are mutually exclusive but they are not independent because if you know one of them occurred you know 100% for sure the other did not. Independent means knowing one tells you nothing about the probability of the other one. So mutually exclusive and independent are very different. Jan 1, 2016 at 19:46
• An example of two events that are independent but not mutually exclusive is as follows. Suppose you roll a die once. Let $A$ be the event that the number is in $\{1,2,3,4\}$ and let $B$ be the event that the number is even. Then knowing $A$ occurred doesn't change the fact that $P(B)=1/2$. That would be written as $P(B)=P(B|A)$ Jan 1, 2016 at 19:49