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 $0.2$

Please explain

  • $\begingroup$ Subtract $P(C\cap D^c)$ rather than $P(C)$. $\endgroup$ – John Dawkins Jan 1 '16 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)

  • $\begingroup$ When calculating union of C and D, i.e. $P(C∪D)$, why are you subtracting the intersection of C and D, i.e. $P(C∩D)$ $\endgroup$ – Rhonda Jan 1 '16 at 20:46
  • 1
    $\begingroup$ 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 $P(C and D) = P(C\cap D) = P(C).P(D)$. $\endgroup$ – Satish Ramanathan Jan 1 '16 at 21:02
  • $\begingroup$ You are welcome!! $\endgroup$ – Satish Ramanathan Jan 1 '16 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.

  • $\begingroup$ How is mutually exclusive different from independent. Please explain. $\endgroup$ – Rhonda Jan 1 '16 at 19:44
  • 3
    $\begingroup$ 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. $\endgroup$ – Gregory Grant Jan 1 '16 at 19:46
  • 1
    $\begingroup$ 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 $P(B)=P(B|A)$ $\endgroup$ – Gregory Grant Jan 1 '16 at 19:49

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