GMAT Quant Review 2nd Ed.-Discrete Probability Example

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ananthbi: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Thu Jul 12, 2012 4:38 am; Thanked: 1 times

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 Aâˆ©C is a subset of A, and that P(A or C) >= P(C) = 0.85 since C is a subset of AâˆªC. 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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Source: Beat The GMAT — Problem Solving | Wed Feb 06, 2013 5:58 am

HerrGrau: Senior | Next Rank: 100 Posts; Posts: 31; Joined: Sun Jan 27, 2013 9:53 amFollowed by: 2 members; Thanked: 6 times; GMAT Score:770

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.

"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land

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ananthbi: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Thu Jul 12, 2012 4:38 am; Thanked: 1 times

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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Tommy Wallach: GMAT Instructor; Posts: 451; Joined: Thu Jan 21, 2010 11:58 amFollowed by: 120 members; Location: New York City; Thanked: 188 times; GMAT Score:770

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

Tommy Wallach, Company Expert
ManhattanGMAT

If you found this posting mega-helpful, feel free to thank and/or follow me!

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ananthbi: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Thu Jul 12, 2012 4:38 am; Thanked: 1 times

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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