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## Why is it always 1-(Probability of it NOT Occurring)?

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### Why is it always 1-(Probability of it NOT Occurring)?

by GTwinkie » Sat Dec 06, 2008 9:56 pm
A machine consists of three components, which are components X, Y, and Z. These three components operate properly independently of one another. The machine operates properly only when component X operates properly and at least one of components Y and Z operates properly. The probability that component X operates properly is 0.8, the probability that component Y operates properly is 0.4, and the probability that component Z operates properly is 0.3. What is the probability that the machine operates properly?

A) 0.096
B) 0.252
C) 0.464
D) 0.583
E) 0.648

In all the Kaplan explanations for probability questions, it always says, "It is easier for us to find the probability of XYZ not occurring".

Can anyone explain why that is? Why can't we just find the probability of XYZ actually occurring?

For example, in the question above, they tell us outwardly the probability of Components Y and Z operating properly, 0.4 and 0.3 respectively. My inclination is to simply multiply the probability of Component X working (0.8) with the probabilities of Y and Z working, 0.8 * 0.4 * 0.3 = 0.096. While, I know that sounds unreasonably low and is ultimately the incorrect answer, why doesn't this simple approach work?

The explanation states that you must first find the probability of Y and Z NOT operating properly, i.e.

1 - 0.4 = 0.6
1 - 0.3 = 0.7

Then multiply them to get 0.42 and subtract it from 1 to find the probability of either Y or Z operating properly, which is 0.58. At that point you just multiply 0.58 by 0.8 and the answer is 0.464.

If anyone can shed light on this, I'd appreciate it. Thanks everyone!

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by Neo2000 » Sun Dec 07, 2008 11:00 am
The question states that "X and atleast one of Y or Z"

You therefore have a number of situtations

X works,Y doesnt, Z does
X Works, Y does, Z doesnt
X works, Y works, Z works

What you are doing is considering only the third situation without considering the first 2

Since we know that total probability = 1
Finding out the sum of cases that dont work and subtracting that from 1 gives you the probability that does work

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by Jen Kedro » Wed Dec 10, 2008 12:41 pm
Exactly...since the probability of something happening, and the probability of that same thing NOT happening, add to 1 (i.e. if probability it will rain is 30%, probability it won't rain is 70%, or total of 100%), then you can use that relationship if it seems like calculating what you are asked for is complicated or very time-consuming....instead, if it's easier and quicker to find the probability that something will NOT happen, you can try that route and then just subtract from one. It's not that you HAVE to use that approach, but it is generally much faster for questions on the GMAT where time is already an issue.

Note that this is pretty common on questions that ask you to find the probability of "at least"....xyz...when you see that phrase, consider this approach to the problem.
Jen Kedrowski