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**CFA Practice Question**

If the stock market has a 65% probability of going up on any given day, what is the probability that one will observe at least 3 up days in a week?

A. 23.5%

B. 42.3%

C. 76.5%

**Explanation:**Since there are only two possible outcomes, up or down, we use the binomial distribution.

Prob(At least three up days) = 1 - Prob(Zero up day) - Prob(One up day) - Prob(Two up days) = 1 -

_{5}C

_{0}x (0.65)

^{0}x (0.35)

^{5}-

_{5}C

_{1}x 0.65 x (0.35)

^{4}-

_{5}C

_{2}x (0.65)

^{2}x (0.35)

^{3}= 1 - 0.0053 - 0.0488 - 0.1812 = 0.765, or 76.5%

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**User Contributed Comments**
19

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

Bibhu |
5 has been used because market works for 5 days in a week only. |

uberstyle |
The 5 days got me. Glad the answer assuming 7 wasn't a choice, as I'm sure it will be on exam day... |

Hope02 |
Got me too. If you sure it's gonna be on the exam, I'd better revise it... |

cahiz84 |
I thought it was .65x.65x.65 Why the 1st day is 0 and not 1? |

s16liu |
it's for 0 day up within 5 biz days. |

StanleyMo |
lol, tricky questions |

mattg |
Another way to look at this one: 65% chance of going up is greater than half, three days in a week is just over half, only answer above 50% is C |

Bobokoko |
I've studied all 4200 questions in the Schweser q-bank and by FAR, analyst notes have the hardest questions. It's not even close. |

wolf2020 |
is there also a quick way how to calc bin. dist with TI? thanks in advance |

trikee01 |
if you think of this question logically, you can save a lot of time. If there is a 65% chance of the stock going up each day, and they want the odds of 3 out of 7 days which is less than 50% of the days in a week, then the answer has to be greater than 65%. |

Bretton |
Isn't prob. of one up day equal to [(.35)^4)*(.65)] giving us .0098? |

harpalani |
very cool, trikee01!! |

indrayudha |
actually trikee01, the odds of 3 out of 5 market days, hence more than 50%. |

dagibbo |
Why is it "1-(the formula)??? Shouldn't this use the binomial distribution formula, which is 5C3*(.65)^3 *(.35)^2??? |

mpeterson |
What about after hours? |

ashish100 |
HOLY EFFING SHIT. I FINALLY GOT THIS!! I'll try to explain below. I was trying the same thing as dagibbo. (for a looongg time) |

ashish100 |
The thing that makes this problem so tricky is that the question is asking "AT LEAST" 3 days, not 3 days. If it was asking us EXACTLY three up days, then we can use the formula on BAII: (5 2nd Ncr 3) * (.65)^3 * (.35)^2 = .3364 But its not. So we have to find the probability of 4 up days and 5 up days also. (Because both of them can also be considered "at least" 3 up days.) So final answer is summing up the probabilities of getting exactly 3 up days, 4 up days, and 5 up days. 4 up days = (5 2nd Ncr 4) * (.65)^4 (.35)^1 = .3124 5 up days = (5 2nd Ncr 5) * (.65)^5 * (.35)^0 = .1160 .3364 + .3124 + .1160 = .7647 or 76.5% |

atlootah |
ashish thanks! |

clafuente |
still thinks in they are uing binomial bc they say market can go up or dwon, but what if there is no change from previous day.... is that matter for takiing into account? |