- CFA Exams
- CFA Level I Exam
- Study Session 2. Quantitative Methods (1)
- Reading 8. Probability Concepts
- Subject 2. Unconditional, Conditional, and Joint Probabilities

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**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.

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**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 |