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**CFA Practice Question**

An economist predicts that over the next year, there is a 60% probability that oil prices will fall slightly and a 20% probability that new estate tax legislation will be enacted. According to this prediction, the probability that at least one of these two events will occur is closest to ______.

A. 12%.

B. 80%.

C. 68%.

**Explanation:**The probability that one of these events will occur: 1 - (1-60%) x (1- 20%) = 68%

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**User Contributed Comments**
16

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

Miketo |
Or Probability of A 60% plus Prob of B 20% equals 80% minus prob of both occuring which is .60 x .20- .12 so 80% - 12%= 68% |

daveyd83 |
The question asks the probability that "at least one event occurs" not "only one event occurs." Therefore, the probability of both happening should not have to be subtracted. Both would be "at least one" |

Kashi2010 |
Surely wrong (again!).. The probability that ONLY one event will occur is 68%, the probability that AT LEAST ONE is 80%..? |

Merke |
Kashi2010, 68% prob that any of these two events will occur, it doesn't matter which one since they're independent. So, you need to take joint probability which is 60%+20%, but subtract the overlapping probability which is 60%*20%. So you get 0.6+0.2-(0.6*0.2) = 0.68. |

arudkov |
2Merke - AT LEAST one event, in my opinion, means P(AorB)+P(A&B) |

Clude |
P(a or b)=P(a)+ P(b) - P(a and b) = 60%+20%-60%*20% |

Clude |
AT LEAST one happens = A happen; B happen; Both happen. |

PattyO |
Think of it as: 1 - the probability that neither happens. |

Mikehuynh |
P(A or B) = P(A) + P(B) - P(A and B) = .6 + .2 - .6*.2 = .68 |

SeanDec10 |
You subtract P(A and B) because you are double counting when you add P(A) and P(B). i.e. the probability P(A) also includes instances where B occurred too, so to remove the double count, you use: P(A) + P(B) - P(A and B) |

SeanDec10 |
It is best to think of it like PattyO said above though |

CalebMast |
P(A) + P(B) - P(A and B) is how I would have solved, but it seems that PattyO's logic is what gets the given answer of 1 - (1-.6)*(1-.2), and I think this is fair, given that at least one happening would be the same as OR, for there are only 2 options. I haven't done the logic proofs (that I wouldn't know anyway) for more than one event, such as P(A or B or C) and at least one happening from that set, but it seems that the 1 -(1-A)*(1-B)*(1-C) would flow better than any equation that would involve removing the joint probability, as compared to the removal of the each probability adjusted individually (1-A,B,or C) then removed together in that fashion. The OR probability value (.68 here) should be different when more variables are added, but it happened to be the same because there are only two possibilities. I'm definitely not the philosopher or mathematician, but hopefully my long response will help someone. |

jjhigdon |
To the comment that the probability of at least one happening means one or both, so probability of both shouldn't be subtracted - First part right, second part wrong. The probability that A happens includes A happening by itself as well as instances of A happening when B also happens. Same goes for P(B). Therefore, P(AB) needs to be subtracted to avoid double counting the instances in which both happen (since its included in P(a)and P(b)). |

GBolt93 |
So much simpler to think of at least one as 1- probability of neither. 1-(.8x.4) |

Rsanches |
Don't confuse the question, we only need to calculate the probability that both doesn't happen: =1-(.4x.8) = 68% |

tmaetzold |
To simplify the problem, you can do a binomial tree and identify the outcomes whose probability you should add up. Just a thought in case you're struggling with the setup. |