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GMAT Quant Review 2nd Ed.-Discrete Probability Example

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GMAT Quant Review 2nd Ed.-Discrete Probability Example

by ananthbi » Wed Feb 06, 2013 5:58 am
Consider an experiment with events A, B, and C for which P(A) = 0.23, P(B) = 0.40, and P(C) = 0.85. Also, suppose that events A and B are mutually exclusive and events B and C are independent. Then

Step 1: P(A or B) = P(A)+P(B) (Since A and B are mutually exclusive)
= 0.23+0.40
= 0.63
Step 2: P(B or C) = P(B)+P(C)-P(B)P(C) (by independence)
= 0.40+0.85-(0.40)(0.85)
= 0.91
Step 3: P(A or C) and P(A and C) cannot be determined using the information given. But it can be determined that A and C are "not mutually exclusive" since P(A)+P(C)=1.08, which is greater than 1, and therefore cannot equal P(A or C); from this it follows that P(A and C) >= 0.08. Once can also deduce that P(A and C) <= P(A) = 0.23, since A∩C is a subset of A, and that P(A or C) >= P(C) = 0.85 since C is a subset of A∪C. Thus, one can conclude that 0.85 <= P(A or C) <= 1 and 0.08 <= P(A and C) <= 0.23

I do understand Step 1 and Step 2 but Step 3 I have difficulties understanding it. I also understand the part that A and C are "not mutually exclusive" but the rest of the statement beginning from P(A and C)>=0.08, I fail to comprehend. I tried hard but I cannot really understand that explanation.

Can someone please throw some light on this? I am just beginning to familiarize myself with Discrete Probability (which I am extremely bad at).

Source: GMAT Quant Review 2nd Edition
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by HerrGrau » Wed Feb 06, 2013 12:04 pm
Hi,

I took a quick look at the Quant Review second edition but I couldn't find this question. Maybe I missed it. What number is it?

HG.
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by ananthbi » Wed Feb 06, 2013 1:57 pm
Hi,

This is a part of the example for Discrete Probability (Page 30) in the 2nd Edition. I was working my way through it and I was stumped by this example. Would appreciate if someone could explain it. Thanks again!
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by Tommy Wallach » Wed Feb 06, 2013 9:10 pm
Hey Anant,

Step 3 is made up of a few discrete parts.

Part 1: If the individual probabilities of A and C add up to something more than 1, then A and C can't be mutually exclusive. If they were, there'd be a 1.08 chance of getting (A or C) (because, in an OR situation, you always add up the individual probabilities), which is impossible.

Part 2: Because we know that A and C aren't mutually exclusive, we know there's at least some chance that they can happen at the same time. This chance has to be greater than .08 (the amount that the sum was bigger than 1). Why? Just because of the logic of probabilities. Let's look at an example, to be sure:

If I have a 75% chance of getting Prize A, and a 75% chance of getting Prize B, my odds of getting both will be (.75 * .75) = .5625. Notice that if I add up 75 and 75, I get 150. That's 50 more than 100%, yet my actual chance is 56%, which is bigger than 50. This will always be true. You can simply take it as a rule.

Part 3: Obviously the probability of two things (both of which have an individual probability of less than 100%) happening at the same time will always be less than the probability of just one of them happening (because when you need two things to happen, you multiply their probabilities together).

Part 4: By the same logic, the probability of one of two things (both of which have an individual probability of more than 0%) happening will always be bigger than the probability of only one of them happening. So P(A or C) must bigger than P(C).

Part 5: So P(A or C) must be between .85 and 1. And P(A and C) must be between .08 and .23 (The probability of just P(A)).

Did that help at all? : )

-t
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by ananthbi » Thu Feb 07, 2013 9:06 am
Thanks Tom! That was perfect and cleared a lot of questions I had. Appreciate the help!
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