- CFA Exams
- CFA Level I Exam
- Study Session 2. Quantitative Methods (1)
- Reading 8. Probability Concepts
- Subject 1. Introduction

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**CFA Practice Question**

Suppose that only three types of birds frequent your neighborhood and for a four-hour period, you record the birds you observe flying into your backyard. During that time, you observe 19 cardinals, 16 blue jays, and 12 robins. If each bird is equally likely to fly into your backyard, what is the probability that the next bird you observe is a robin?

Correct Answer: 0.255

p = 12 / (19 + 16 + 12) = 0.255

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**User Contributed Comments**
19

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

leoo |
how? shouldn't it be 1/3 sine each bird is equally likely to fly there |

KD101 |
THE ANSWER IS 1/3 - as it says that PROB(NEXT BIRD YOU OBSERVE IS ROBIN) - Think of this as: You toss a coin 20 times - you get 15 Heads and 5 Tails - still Prob(Tail on next toss) = 0.5 |

Gina |
leoo, this is a relative frequency. since there are fewer robins than cardinals, your probability to see a robin is lower than the one to see a cardinal (the latter would be p(Cardinal)=.404 |

jminard |
agree with Gina. the answer is correct. |

Arron |
impirical probability from historical data. |

creativemny |
The question implies conflicting methods but tells you each bird is equally likely which, if you ask me, takes empirical off the table. I would answer 1/3. |

stefdunk |
the question does not say that each type of bird is equally likely, but each individual bird, which have a 19:16:12 proportion. |

mtcfa |
It says that during the 4 hour period, you witnessed x amount of each bird. It never says there are in fact more cardinals etc, just the observation during this one 4-hour trial. |

JP09 |
They key is that it says each bird is equally likely to fly into your backyard, not each type of bird. This small difference is what leads to the use of historical/observational probabilities being used as opposed to a simple 1/3. |

mtcfa |
I understand now. Good explanation JP09. |

surob |
I think the credit should go to stefdunk. But really good explanation from both of you. Thanks |

Bibhu |
Good explanation JP. |

viannie |
equally likely refers to randomness ... I think ;) but it's probability is based on empirical as suggested by Gina etc |

rm001 |
the question says each bird is equally likely to 'fly' but it doesn't imply that the number of all birds is equal....answer is correct |

LionHero |
Agree with JP09 |

gmilchev |
Equally likely to FLY in is 1/3 but the probability that it is a Robin is 25.5% |

msk500 |
The numerator = the number of birds that meet 'Robin' outcome (ie 12). The denominator = the number of birds that meet 'Bird' outcome, ie 47. Thus, 12/47 = 0.25532. |

dbedford |
Remember: P(A) = a/(a+b); therefore, a = # Robins (12) and b = # of Cardinals (19) and # of Bluejays (16). So P(Robin) = 12/(12+19+16) |

khalifa92 |
this is empirical probability equally likely here is to confuse you, people, lets say there's a box with 10 balls( Red=8, Y=1, B=1) they are equally likely to be picked up but still it wont be 1/3. |