- CFA Exams
- CFA Level I Exam
- Study Session 2. Quantitative Methods (1)
- Reading 8. Probability Concepts
- Subject 3. Addition Rule for Probabilities: the Probability that at Least One of Two Events Will Occur

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**CFA Practice Question**

In a town, 36% of households own a dog, 20% own a cat, and 60% own neither a dog nor a cat. If we select a household at random, the chance that it owns both a dog and a cat is ______.

Correct Answer: 0.16

The general addition theorem says P(A or B) = P(A) + P(B) - P(A and B), or equivalently,

P(A and B) = P(A) + P(B) - P(A or B) = 0.36 + 0.2 - 0.4 = 0.16.

Let A be the event that the selected household owns a dog and B be the event that the selected household owns a cat. The event A or B indicates the selected household owns either a dog or a cat. (P(A or B) = P(own either a dog or a cat) = 1 - P(own neither a dog nor a cat) = 1 - 0.6 = 0.4.

The general addition theorem says P(A or B) = P(A) + P(B) - P(A and B), or equivalently,

P(A and B) = P(A) + P(B) - P(A or B) = 0.36 + 0.2 - 0.4 = 0.16.

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**User Contributed Comments**
19

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

gruszewski |
why not 0.36x0.20 ? |

Gina |
i am also puzzled by the answer. if you do (.36)*(.20)=.072 you arrive at the likelihood that a household has a cat and a dog. you still need to relate it to the number of households that have a pet. in other words, how many of the 40% of households that have either a cat or a dog do have both, ie 7.2% have both, 40% have one or the other. 40%x=7.2% x=18% = households that have both. what do you think?? |

Hamma |
I am puzzled by wording and math interpretation for it. If 60% of people owns no cat and dog. Then would not 40% people could own a dog or a cat, or both? so 40% is not P(own either a dog or a cat). 40% covers more than as the probability states. Anyone else comment on this? |

Tony1973 |
Let's put it this way: There are 100 households. We number them from 1 - 100. any household in (1 - 60) owns neither a dog or a cat. Then any household in (61 - 100) must own a dog only, or a cat only, or both. Suppose any household in (61 - 80) owns a cat (it may own a dog as well): that's 20%. Then any household in (81 - 100) must own a dog. However as 36 households own a dog, then the rest of 16 households must in (61 - 80) who also own cats. Therefore, 16% own both a cat and a dog. |

Abde |
That was a good explanation Tony! |

lzz326 |
be simple,.36-.20 |

db28luke |
lzz326, I don't believe your solution will always work. |

itconcepts |
Note look carefully at the use of (and) and (or) in the theorem, in the equivalent formula it is swopped around |

o123 |
* P(nD or nC) = .60 therefore: P(D or C) = .40 .40 = .36 + .2 - P(DC) P(DC) = .16 |

maciejf |
If "60% own neither a dog nor a cat." it means 40% has a dog or cat => P(A or B). We are looking for P(AB) so we use basic formula P(A or B)=P(A) + P(B) - P(AB) P(AB)=P(A) + P(B) - P(A or B) P(AB)=0,36 + 0,20 - 0,40 P(AB)=0,16 |

ilgibe |
P(Cat&Dog)is a probability dependending on P(HavePet). Otherwise P(Cat&Dog) would have been equal to P(Dog)*P(Cat). |

jpducros |
I suggest to draw a "potatoe" quick drawing to see immediately that P(A and B) = P(a)+p(B)-(1-P(nA or nB)). |

DariSH |
easy way: put together cat and dogs owning households and petless households: 36%+20%+60%=116%. The total amount of households should be 100%, therefore extra 16% are those, having both cats and dogs. |

LoveIvie |
Good explanation o123! |

bidisha |
very nice 0123 |

tichas |
Hypothetical example 100 is comprised of cats and dogs 36% = 36 dogs 20%= cats 60% = neither cats nor dogs Total = 116% and probability has to be 1 0r 100% but in this case we have 116% which means 116-100 is our answer which is 16%. |

dservan |
my problem is that i dont know when to use what formula. why not use P(AB) = P(AIB) P(B) or P(AB) = P(A) + P(B) ? |

maryprz14 |
20+36-X = 40 >>>>>> X=16 :) |

CFAJ |
This question doesn't specify whether these events are independent or not |