- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 4. Common Probability Distributions
- Subject 5. Binomial Distribution

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**CFA Practice Question**

The probability of an accident-free day at MRC is 60%. Assuming the accident-free days at MRC are independent, for a random sample of 15 days, the probability of less than 14 accident-free days is ______ (to the nearest 0.1%).

B. 99.5%

C. 0.5%

A. 4.7%

B. 99.5%

C. 0.5%

Correct Answer: B

Let X count the number of accident-free days. The events are independent with only two outcomes: an accident-free day or a day that included an accident. Binomial random variable x has N = 15 and p = 0.6. Computing the binomial distribution for binomial random variable X (using the calculator), we get P(X < 14) = 1 - P(X >= 14) = 1 - [p(14) + p(15)] = 1 - [0.0047 + 0.0004] = 1 - 0.0051 = 99.5% (to the nearest 0.1%).

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**User Contributed Comments**
11

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

mtcfa |
Howd o you get P(14) and P(15) to equal .0047 and .0004, respectively? |

Guerra |
P(14)=(15!/(14!x1!))x 0,6^14 x 0,4^1 P(15)=(15!/(15!x0!))x 0,6^15 x 0,4^0 |

tenny45 |
If you just look at the choices, B is the only one that seems logical. No calculation actually needed. |

DAS11 |
Why are we including p(14) if it is asking for P(X<14)? |

DAS11 |
nevermind..got it. |

euniceyew |
is this NCr since the order is not important? |

Kobe8kenji |
yes this is nCr. In binomial it is a must to use nCr. |

johntan1979 |
99.482797% |

ioanaN |
p(less than 14 acc)=1-p(14 acc)-p(15 acc). so in the binomial shouldn't it be 0.4^14X0.6 and 0.4^15? since B(15, p of accident=0.4). Thanks |

mmccoy |
15nCr14 * (.6^14)(.4^1)+1 = .9952 |

MathLoser |
It's weird that I can still receive the same answer if I use: 1 - P(14) P(14) = 15C14 * (0.6^14) x (0.4^1) = 0.00470185 1 - 0.00470185 = 0.99529815 = 99.5% |