- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 3. Probability Concepts
- Subject 2. Unconditional, Conditional, and Joint Probabilities
CFA Practice Question
Roll a die and flip a coin, P(5 given heads) = ______.
B. 1/6
C. 1/8
A. 1/2
B. 1/6
C. 1/8
Correct Answer: B
Because these events are independent events (what happens with the coin has no effect on what happens with the die), P(5 | heads) = P(5). Now, roll a die P(5) = 1/6. So, P(5 | heads) = 1/6.
User Contributed Comments 12
| User | Comment |
|---|---|
| leoo | how, i thought the question is what is the probabilty of rolling a 5 + head on the flip , shouldn't it be 1/6 * 1/2? |
| cfaman | The questions asks the conditional probability. As the two events are independent, you can discard the condition so it's simply P(5) = 1/6. |
| Pooh | cfaman, I believe you meant to say unconditional probability? |
| yanpz | To Pooh, no, it's conditional probability. Actually for this question rolling a die won't affect the probability of flip coin, so the conditional probability value is the same as unconditional probability for given rolling die number, flip a coin. |
| tssverma | tricky question. The question implicitly expects the reader to understand the coin tossing and rolling a die that they are mutually exclusive. Normally questions are more clear. |
| PeterW2006 | P(5 and Heads) = P(5 | Heads) x P(Heads) P(5 and Heads) = P(5) x P(Heads) = 1/6 x 1/2 = 1/12 P(Heads) = 1/2 1/12 = P(5 | Heads) x 1/2 P(5 | Heads) = 1/6 |
| mtcfa | Even if you work out the whole problem utilizing the joint probability formula [ P(5!H) = P(5H)/P(H)], you still get 1/6. But if you can just see that they are independent events, you know the answer. |
| faya | mtcfa: But P(5H)=1/6=0.1667 and P(H)=0.5 So P(5H)/P(5)=0.1667/0.5=0.3333 which is not equal to 0.1667. If they say P(5 given Heads) are we not to assume conditional prob - I'm confused. |
| adenisov | PeterW2006: without calculations P(5 and Heads) = P(5 | Heads) x P(Heads) P(5 and Heads) = P(5) x P(Heads) so P(5 | Heads) x P(Heads) = P(5) x P(Heads) and P(5 | Heads) = P(5) |
| StanleyMo | i believe given is the key word, when talking about given, you need to determine yourself whether they are "dependant" or "independant". |
| maciejf | Given is a very key word, we know that one event happend, so we have to count only probablility of rolling a die and having 5. |
| rsanfo | Even if you think it is conditional, you get the same answer: P(5|H) = P(5H) / P(H) = 1/12 divided by 1/2 = 1/6 |