# Probability problem

• Jun 19th 2010, 02:04 AM
cloud5
Probability problem
The probabilities that machines X, Y, Z will be performing well are $\frac{1}{3},\frac{1}{4},\frac{1}{5}$. Find the probability that at most two of the machines will be operating.

P(at most two operating)
=1-P(all operating)
= $\frac{59}{60}$

This is found in my math book. Can anyone explain why we need to use 1-P(all operating)? and I found out this 1-P(all operating) cannot be used when finding P(all machines not working)... Why?
• Jun 19th 2010, 02:39 AM
undefined
Quote:

Originally Posted by cloud5
The probabilities that machines X, Y, Z will be performing well are $\frac{1}{3},\frac{1}{4},\frac{1}{5}$. Find the probability that at most two of the machines will be operating.

P(at most two operating)
=1-P(all operating)
= $\frac{59}{60}$

This is found in my math book. Can anyone explain why we need to use 1-P(all operating)? and I found out this 1-P(all operating) cannot be used when finding P(all machines not working)... Why?

Consider events A,B defined as: A = (at most 2 functioning) and B = (all 3 functioning). A and B are complements of each other. In other words, B means: not A. So P(A) + P(B) = 1.

P(all machines not working) is (1-1/3)(1-1/4)(1-1/5). Do you see why?

Edit: Originally I used X,Y in place of A,B not noticing that this choice of letters conflicted with the machine names.
• Jun 19th 2010, 05:47 AM
cloud5
I understand now. Thank you. :)

Quote:

Originally Posted by undefined
Consider events A,B defined as: A = (at most 2 functioning) and B = (all 3 functioning). A and B are complements of each other. In other words, B means: not A. So P(A) + P(B) = 1.

P(all machines not working) is (1-1/3)(1-1/4)(1-1/5). Do you see why?

Edit: Originally I used X,Y in place of A,B not noticing that this choice of letters conflicted with the machine names.