### CFA Practice Question

There are 434 practice questions for this study session.

### 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%).

A. 4.7%
B. 99.5%
C. 0.5%

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%).

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%