CFA Practice Question

There are 410 practice questions for this study session.

CFA Practice Question

A six-sided die is rolled once and the outcome is observed. Define the events A and B as follows:

Event A: Number observed is even
Event B: Number observed is greater than 3

Find P(A or B).
Correct Answer: 2/3

P(A) = 0.5, P(B) = 0.5, and P(A and B) = P(4 or 6) = 1/3. Then P(A or B) = P(A) + P(B) - P(A and B) = 2/3.

User Contributed Comments 11

User Comment
10klub Can some body explain this a little more for me
Carter Okay. Here we go.

Outcomes for Event A: 2, 4, 6 => 3/6 possible = 1/2.
Outcomes for Event B: 4, 5, 6 => 3/6 possible = 1/2.
Outcomes for A and B: 4, 6 => 2/6 or 1/3.

Thus, we add when A happens (1/2) to when B happens (1/2) but subtract off when they both happen (1/3) because we've already accounted for that. The probability of event A occuring (1/2) includes when B occurs and when B does not occur. That's where it is already accounted for.
aakash1108 @ Carter. Thanks for the explanation.
rsanfo For small number of outcomes, you can just think about it logically without using any formulae:
Is it even or greater than 3?
1 - N
2 - Y
3 - N
4 - Y
5 - Y
6 - Y
arv333 Agree with both rsanfo & Carter. Just dont understand one thing. The P(AB) is 1/3 (4,6 =>2/6 or 1/3). While if I were to calculate using the formula P(AB) = P(A) x P(B), I get 1/4 and not 1/3.
P(A) = 1/2
P(B) = 1/2
P(AB) = 1/2 x 1/2 = 1/4

Im still lost ???? Can someone clarify?
elvinos arv333, A and B are not independent events since P(A|B)is not eqal to P(A). (Just think, the probability of getting even number of 6 is 1/2, but the probability of getting even number of 3 isn't 1/2).
That's why your rule doesn't work.
johntan1979 Don't make the mistake P(AB) = .5 x .5

For A AND B to be true, 2 and 5 cannot be included, because 2 is less than 3 (event B) and 5 is not even (event A)
Amrokken great explanation Carter
assiduous Initially I was still unclear just as arv333. Without paying attention to the "die is rolled once" part both events do read as independent events. It would be helpful if Event B read "number observed in Event A is greater than 3."
idzani Good question. Totally overlooked the fact that P(A and B) is the common outcomes shared by P(A) and P(B) which means having the die showing 4 & 6 only.
Frontier17 P(AB) has to be even and bigger than 3. Only 4 and 6 can be included. Hence 2/6
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