- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 3. Probability Concepts
- Subject 2. Unconditional, Conditional, and Joint Probabilities

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**CFA Practice Question**

For the tree diagram shown below:

B. 2/3

C. 1/3

P(not B | A) = ______

A. 0.1

B. 2/3

C. 1/3

Correct Answer: C

From the tree diagram, we know that 0.3 times P(B | A) = 0.2. Now, 0.3x = 0.2 yields x = 0.2/0.3 = 2/3. Thus, P(B | A) = 2/3. We have: P(B | A) + P(not B | A) = 1. Finally, 2/3 + P(not B | A) = 1 and P(not B | A) = 1 - 2/3 = 1/3.

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**User Contributed Comments**
17

User |
Comment |
---|---|

stevelaz |
Can someone explain this? |

BayAreaPablo |
I agree how can P(B|A) go from .2 to .67? |

BayAreaPablo |
I get it: P(B|A)x P(A)=P(AB) therefore, P(B|A)=? P(A)=.3 P(AB)=.2 Therefore: P(B|A)x.3=.2 P(B|A)=2/3 Now that you have this:Remember P(B|A) + P(not B|A)=1 Now solve for P(not B|A) |

surob |
I don't still understand why 0.2 = P(AB)? |

cp24 |
What's misleading is that probability 0.2 is not P(B/A) but P(BA). |

epizi |
I don't think there is any problem,because the tree indicates that A happens before B, not both happening same time.Therefore P(A)=0.3 and P(B)=? But the P(AB)=0.2 P(AB)=P(A)xP(B/A) 0.2=0.3xP(B/A) P(B/A)=2/3 P(B/A) and P(notB/A) are exhaustive ie sum up to 1 So 1-P(B/A)=P(not B/A) Hence 1-2/3= 1/3 Hope this makes it clear, I did not get it from first try either |

StanleyMo |
thanks epizi. nice explain. |

JKiro |
I find this a bit more clearer... what we know: p(A) * p(B|A) = 0.2; we also know p(A) = 0.3 solving for p(B|A): 0.2/0.3 = 0.6667 what we also know: p(B|A) + p(not B|A) must equal 1 we have calculated p(B|A) solving for p(not B|A): 1 - 0.6667 = 0.3333 |

fedha |
JKiro I am not sure how you know p(B|A)is 0.2?? I hope you meant p(BA)= 0.2 Here is how would solve this problem. Epizi is right in solving the problem P(B|A) = P(BA)/P(A) P(B|A) = 0.2/0.3 P((B|A) = 0.667 We know that P(B|A) + P(Not B|A) = 1 We already solved for P(B|A) = 0.0667 Therefore P(Not B|A) = 1-0.667 = 0.333 ~ 1/3 |

loisliu88 |
I don't understand how can P(B/A)+P(notB/A)=1, what about P(notA). shouldn't it be P(B/A)+P(notB/A)+P(notA/B)+P(notA/notB)=1? plz explain it to me. |

harpalani |
Good explanation epizi!! |

LordRommel |
Oh. It is simple. 0.01/0.03 = 1/3 |

Saxonomy |
If I understood the diagram, the answer would be easy. Loisliu88, too much going on in your comment: P(B|A) + P(notB|A) = 1 Ignore the "|A", just know that B + "notB" will always be 1. Prob that LeBron wins the title (given the fact that the Lakers/Celtics are out) + the prob that LeBron does not win the title (given the fact that the Lakers/Celtics are out) equals 1. |

2014 |
Good explnation Fedha |

sgossett86 |
CFAs, this problem upset me. Like they say above, turns out I wasn't reading the chart right. .2=P(AB) .3=P(A) P(A|B)=P(AB)/P(A)=.2/.3 P(A|B)+P(A|notB)=1 .2/.3+P(A|notB)=1 P(A|notB)=1/3 |

Scc0813 |
Why is 0.2 P(AB) and not P(B|A)? |

Safiya921 |
I didn't read the Tree Diagram correct. I thought P(B I A) = 0.2, then I got to know from your comments that it's P (AB) = 0.2. Thank you all. |