Simple Probabilty Question (That I am struggling with!)

T

tfs985

New member
Joined
Apr 17, 2011
Messages
13
Hi guys, I have a (seemingly) simple probability question that I am struggling to figure out. It goes like this:

If you set 2 alarm clocks, and there is a 90% chance that either alarm clock will wake you up, what is the probability that you will oversleep?

When I apply the law of addition and the law of addition for mutually exclusive events, I get a number larger than 1.

Any help would be greatly appreciated!

Thanks
 
tfs985 said:
Hi guys, I have a (seemingly) simple probability question that I am struggling to figure out. It goes like this:
If you set 2 alarm clocks, and there is a 90% chance that either alarm clock will wake you up, what is the probability that you will oversleep?
Click to expand...

\(\displaystyle \mathcal{P}(C_1\cup C_2)=\mathcal{P}(C_1)+\mathcal{P}(C_2)-\mathcal{P}(C_1\cap C_2)\)

OR \(\displaystyle 1-\mathcal{P}(C_1\cup C_2)'\)
 
Binomial Distribution Question

I was looking for more Statistics problems to prepare for an upcoming test. I solved the 2 alarm clock problem with the following

n= 2, p=.90, and figured that q=.10

M= (2)(.90) = 1.8
standard deviation is square root of (2)(.90)(.10) = .424
 
HeidiO said:
I was looking for more Statistics problems to prepare for an upcoming test. I solved the 2 alarm clock problem with the following
n= 2, p=.90, and figured that q=.10
M= (2)(.90) = 1.8
standard deviation is square root of (2)(.90)(.10) = .424
Click to expand...

@HeidiO, please start a new thread with a new question.
You have no right to hijack someone else's thread.

Please read the forum rules before you precede.
 
HeidiO said:
I was looking for more Statistics problems to prepare for an upcoming test. I solved the 2 alarm clock problem with the following

n= 2, p=.90, and figured that q=.10

M= (2)(.90) = 1.8
standard deviation is square root of (2)(.90)(.10) = .424
Click to expand...
So what answer did you get (most of what you wrote has nothing to do with the problem).
 
You must log in or register to reply here.
Top