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GMAT Problem Dissection: Probability Part 2

July 25, 2013 by

Alright, you’ve been studying the GMAT basics. You understand what is really being tested along with the problem types, methods, and strategies. You’ve even memorized all the basic formulas and taken a dive into some of the more challenging content areas. So what’s next? That’s simple, practice…over and over and over again. If you do this, you are likely to start hitting tough probability questions at some point. Along with combinations and permutations, this is a content area that the GMAT can use to up the degree of difficulty quite a bit. So let’s take a look at a tough probability question and break it down:

A fair coin is tossed five times. What is the probability that it lands heads up at least twice?

(A) 1/16

(B) 5/16

(C) 2/5

(D) 13/16

(E) 27/32

The key to solving this is the phrase ‘at least twice.’ This means that … Read full post

GMAT Question Breakdown: Probability

June 11, 2013 by

2 Dice and the GMAT. What’s the relationship?

January 26, 2012 by

While my students certainly accept that they have to take the GMAT, I often hear complaints about the applicability of GMAT topics – especially math – to real world situations.  My usual response to this observation is that the GMAT is using math questions to test critical thinking skills that business schools consider essential.

However, some GMAT problems have surprising real world applications outside the realm of business.  The sample problem below is an example of one such question.  Specifically, it can help you win at craps.

If you want to try the problem on your own, skip down to it now and then return here – part of this discussion will give you clues about the correct answer.

For those of you not familiar with craps, the basic play is fairly straightforward.  You roll two dice and if the sum of the result on each die is 7 … Read full post

Kaplan GMAT Sample Problem: Probability and “at least”

August 29, 2011 by

Probability questions can be among some of the more advanced and trickier problems you’ll face on the GMAT Quantitative section.  Be sure to pay attention to the wording of word problems such as this one; in this case when asked about a scenario with “at least twice”, it will be more efficient to solve for that NOT happening and subtract from 1 (since the probability of something happening plus the probability of that same thing NOT happening should add up to 1, or 100%.)

Question:

A fair coin is tossed five times.  What is the probability that it lands heads up at least twice?

(A) 1/16

(B) 5/16

(C) 2/5

(D) 13/16

(E) 27/32

Solution:

The key phrase to solving this sample GMAT problem is ‘at least twice.’  This means that out of our five flips, two, three, four and five heads are all desired outcomes.  On problems such … Read full post

Kaplan GMAT Sample Problem: Probability Problem Solving

August 24, 2011 by

Try this advanced GMAT probability question, testing your knowledge of the ins and outs of how probability works.

Problem:

The events A and B are independent.  The probability that event A occurs is 0.6, and the probability that at least one of the events A and B occurs is 0.94.  What is the probability that event B occurs?

(A) 0.34

(B) 0.65

(C) 0.72

(D) 0.76

(E) 0.85

 

Solution:

In order to find the probability that event B occurs in this problem, we need to set up and equation that includes the probabilities we are given and allows us to solve for B.  We are told that the probability that at least one of A or B occurring is 0.94.  ‘At least one of A or B’ means that an outcome is desired if A occurs and B does not, B occurs and A does not or A and … Read full post

Kaplan GMAT Sample Problem: Advanced Probability

June 29, 2011 by

Try your hand at this GMAT Problem Solving Probability question.  Remember, on test day, it is easy to get distracted and overwhelmed by more complex word problems, but it’s important to stay calm, and start working through what you DO know, and you’ll often be able to get much further in the question than you might initially think you can.  Good luck!

Sample Question:

Darcy, Gina, Ray and Susan will be the only participants at a meeting.  There will be three soft chairs in the room where the meeting will be held and one hard chair.  No one can bring more chairs into the room.  Darcy and Ray will arrive simultaneously, but Gina and Susan will arrive individually.  The probability that Gina will arrive first is 1/3, and the probability that Susan will arrive first is 1/3.  The probability that Gina will arrive last is 1/3, and the probability that … Read full post

Kaplan GMAT Sample Problem: Probability Problem-Solving

June 9, 2011 by

Try your hand at this sample GMAT problem focusing on a specific probability situation.

Sample Problem:

Each person in Room A is a student, and 1/6 of the students in Room A are seniors.  Each person in Room B is a student, and 5/7 of the students in Room B are seniors.  If 1 student is chosen at random from Room A and 1 student is chosen at random from Room B, what is the probability that exactly 1 of the students chosen is a senior?

 

(A) 5/42

(B) 37/84

(C) 9/14

(D) 16/21

(E) 37/42

 

Solution:

In this problem we are asked to determine the probability of choosing exactly 1 senior.  To be more specific, this means that we will need to select 1 senior and 1 non-senior.

There are two ways this can be done.  We can select a senior from Room A and a non-senior … Read full post

Kaplan GMAT Challenge Problem: Combined Work

February 9, 2011 by

Work problems are definitely not as common on the GMAT as, say, solving simultaneous equations might be; but many test-takers are wary of these problems since they are not as commonly used in everyday life as averages are, for example.  The key to most of these problems, though, is to know the work formula, and how to use it.  Try the challenge problem below for an advanced twist that includes probability along with the work formula.  

Sample Problem:

Mike and Emily need to build 2 identical houses.  Mike, working alone, can build a house in 6 weeks.  Emily, working alone, can build a house in 8 weeks.  To determine who will do the building they will roll a fair six-sided die.  If they roll a 1 or 2, Mike will work alone.  If they roll a 3 or 4, Emily will work alone.  If they roll a 5 or 6, … Read full post

GMAT Probability 101, Part 5

January 24, 2011 by

In our GMAT Probability 101—Part 3 article, we were in the middle of discussing how to find the probability of getting at least two heads when flipping five fair, two-sided coins.  In our Part 4 article, we went over combinations and permutations, which are essential to answering questions such as the one above.  Specifically, we will need to use the combinations formula, n!/(k!(n-k)!)

As we went over last time, to answer this problem most efficiently we will find the probability of NOT getting at least two heads, and subtract this total from one.  This means the desired outcomes are zero heads or one head.

The only way we can get zero heads is to get all tails, so there is one possible outcome for zero heads.

 In order to determine the number of ways we can get exactly one head, we must use the combinations formula, using the … Read full post

GMAT Probability 101, Part 4: Combinations/Permutations

January 19, 2011 by

We have looked at increasingly difficult GMAT probability scenarios in three previous articles.  As these probability questions became more advanced we saw that we would need to master combinations and permutations in order for success on test day.  Here, we will go over the basics of combinations and permutations.

Combinations and permutations problems involve creating groups and arrangements and fall into three categories.  When facing these problems, you should first determine which of these three categories you are in and then apply the appropriate formula.

The first question type will give you a number of items and ask you how many ways they can be arranged.  This is a permutation question.  For example, a problem could ask you how many ways you can order four people.  If we call these people A, B, C and D, one possible arrangement is A first, then B, then C and then D.  … Read full post

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