January 26, 2012
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 or 11 you win. If the sum of the results are 2, 3 or 12 you lose. If you roll any other result the game progress to a second step that we do not need to concern ourselves with for purposes of this discussion.
Someone who is not familiar with probability might look at those rules and think, “this game is not fair – I have three ways to lose, but only two ways to win.” But, by understanding probability we can see that craps is actually the most fair game in the casino.
The key is that every result does not have an equal probability of occurring. To role a 2, the first die must roll a 1 and the second die must also role a 1. Likewise, to roll a 12, both dice must come up with a 6. A three is slightly more likely – the first die can be a 1 and the second a 2 or the first die can be a 2 and the second a 1. This gives us four ways to lose in our initial roll.
Rolling an 11 is just as likely as rolling a 3. You can roll a 5 on the first die and a 6 on the second or a 6 on the first and a 5 on the second. However, we can roll a 7 in many ways. We can roll a 1 first and a 6 second, a 2 first and a 5 second, a 3 first and a 4 second, a 4 first and a 3 second, a 5 first and a 6 second, or a 6 first and a 1 second. Thus, we have six ways to make a seven when we roll our dice.
This means that we have eight ways to win and only four ways to lose. While, the rest of the game pushes the odds back into the casino’s favor, on that initial roll we are twice as likely to win as lose.
This concept of desired versus undesired outcomes will show up on many probability questions on the GMAT – give the problem below a try to see how.
What is the probability of rolling a total of 7 with a single roll of two fair six-sided dice, each with the distinct numbers 1 through 6 on each side?
When confronted with a probability question, always remember two points. First, probability is just desired outcomes divided by possible outcomes, so you will need to find this ratio. Second, if more than one event occurs, “and” means multiply and “or” means add.
In this problem, we want to roll a combined 7. We can accomplish this in a number of ways. We can roll a 1 with our first die and a 6 with our second, a 2 first and a 5 second, a 3 first and a 4 second, a 4 first and a 3 second, a 5 first and a 2 second, or a 6 first and a 1 second.
This gives us a total of 6 desired outcomes. Next, we will need to find our possible outcomes. Since we have 6 possible outcomes for the first die and 6 possible outcomes for the second die we should multiply 6 x 6 = 36 possible outcomes. Also note, that this is an “and” situation – we roll the first AND second die – so we want to multiply rather than add.
Finally, we put desired outcomes over possible outcomes, to get 6/36 as the probability of rolling a combined 7. 6/36 simplifies to 1/6, which is answer (B).
August 29, 2011
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%.)
A fair coin is tossed five times. What is the probability that it lands heads up at least twice?
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 as this one it is important to remember that we can find the probability that something does NOT happen and subtract that from one, in order to find the probability it DOES happen. In this case, if we flip zero heads or one head, we will NOT have at least two heads. Finding these two probabilities, adding them together (remember, that ‘or’ becomes addition in probability) and subtracting from one will be the fastest way to get to the answer.
Because probability means desired outcomes over possible outcomes, we will need to find each of these for this problem. The first outcome we want to consider is zero heads. Only one way exists for this to happen: all of the flips come up as tails. Thus, we have ONE desired outcome in which zero heads appear. The second outcome we are looking for is exactly one head. Five outcomes would provide one head, as the head could be first, with all other flips coming up tails, the head could be second, with all other flips coming up tails, and so on. Therefore, in total we have six outcomes that do not give us at least two heads.
Next, we need to find the number of possible outcomes (our denominator). Each flip has two possible outcomes: heads and tails. In order to find total possible outcomes, we multiply the possible outcomes for each individual flip. As we have five flips, we get 2 x 2 x 2 x 2 x 2 = 32 possible outcomes.
When we put these together we have a 6/32, or 3/16, probability of not flipping at least two heads. Since we found the probability of what we do not want to happen, we still need to subtract our result from one to find the probability it does happen. The math for this is as follows:
1 – 3/16 = 16/16 – 3/16 = 13/16
13/16 is answer choice (D) and is the correct answer.
August 24, 2011
Try this advanced GMAT probability question, testing your knowledge of the ins and outs of how probability works.
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?
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 B both occur.
It is important to remember two rules of probability here. First, when you encounter an ‘or’ situation you add and when you see an ‘and’ situation you multiply. Second, the probability that an event does NOT occur is equal to one minus the probability it does occur.
Based on these rules, we can translate ‘at least one of A or B occurs is 0.94’ to the following equation:
.6B [the probability that both A and B occur] + .6(1-B) [the probability that A occurs and B does not] + B(1-.6) [the probability that B occurs and A does not] = .94
We can simplify this equation to .6B + .6(1-B) + .4B = .94 and solve for B.
.6B + .6(1-B) + .4B = .94
.6B + .6 – .6B + .4B = .94
.6 + .4B = .94
.4B = .34
B = .34/.4
B = 34/40 = .85
Thus, our answer is 0.85, or answer choice (E).
June 29, 2011
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!
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 Susan will arrive last is 1/3. Upon arriving at the meeting, each of the participants will select a soft chair, if one is available. If Darcy and Ray arrive and see only one unoccupied soft chair, they will flip a fair coin to determine who will sit in that chair. By what percent is the probability that Darcy will sit in a soft chair greater than the probability that Gina will sit in a soft chair?
C) 16 2/3%
D) 12 1/2%
First, we need to determine the probability of each of Darcy and Gina sitting in a soft chair. To do so, we must determine all of the possible ways in which the four people can arrive. To visually list these out, we can make the following chart:
Keeping in mind that if Gina arrives first, she will select a soft char, if she arrives second she will select a soft chair and if she arrives third she will be forced to select the hard chair, as all three soft chairs will already have been taken, we know that Gina ends up in a soft chair two-thirds of the time. Thus, Gina has a 2/3 probability of sitting in a soft chair.
If Darcy arrives first she will select a soft chair, if she arrives second, she will select a soft chair and if she arrives third, she will select a soft chair half of the time, as only one will be available and she and Ray will flip a coin to see who gets it. Thus, Darcy only gets a hard chair if she arrives last AND loses the coin flip. The probability of Darcy arriving last is 1/3 and the probability of losing the coin flip is 1/2, so the probability of both of these happening is 1/3 x 1/2 = 1/6. Therefore, the probability of her ending up in a soft chair is 5/6.
Since 5/6 – 2/3 = 1/6, the probability of Darcy sitting in a soft chair is 1/6 more than Gina sitting in a soft chair. At this point it is essential to remember that the question asks for what percent it is greater. In order to determine this, we need to divide the difference of their probabilities by Gina’s probability. (1/6)/(2/3) = 1/6 x 3/2 = 3/12 = 1/4. 1/4 is the equivalent of 25%, which is choice (B) and the correct answer.
Try your hand at this sample GMAT problem focusing on a specific probability situation.
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?
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 from Room B or we can select a non-senior from Room A and a senior from Room B. In probability, ‘and’ means multiply while ‘or’ means add. In this case, we see 2 ‘ands’ indicating that we will multiply probabilities together and one ‘or’ telling us that we will add the resulting products.
Our first option is to get a senior from Room A, which has a probability of 1/6, and a non-senior from Room B, which has a probability of 2/7 (we found this number by taking the probability of selecting a senior in Room B and subtracting it from 1.) Because this is an ‘and’ situation, we multiply 1/6 x 2/7, which equals 2/42.
For our other option we must find the probability of selecting a non-senior from Room A, which has a probability of 5/6, and senior from Room B, which has a probability of 5/7. Again, we have an ‘and’ statement, so we multiply. 5/6 x 5/7 = 25/42.
Lastly, we must address the ‘or’ in our original scenario by adding the two probabilities we determined together. This gives us 2/42 + 25/42 = 27/42. This simplifies to 9/14, which is choice (C).
February 9, 2011
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.
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, they will work together and independently. What is the probability both houses will be completed after 7 weeks?
The first step in solving this problem is to determine how long it would take them to build one house working together. The fastest way to do this is to use the combined work formula, which is AB/(A+B), where A is the time it takes the first person working alone and B is the time it takes the second person working alone. In this problem, the equation gives us (6)(8)/(6+8) = 48/14.
48/14, however, is the amount of time it takes Mike and Emily to build one house together and the problem specifies that they must build two houses. To determine how much time it takes them to build two houses, simply double the time it takes to build one house. Thus, it takes them 2(48/14) = 96/14 weeks to complete both houses. 96/14 expressed as a mixed number is 6 6/7 or 6 weeks and 6 days.
Next, we need to consider the probability component of this problem. We have a 1/3 chance of Mike working alone, a 1/3 chance of Emily working alone and a 1/3 chance of them working together. If Mike works alone, two houses take 12 weeks to build; if Emily works alone, two houses take 16 weeks to build; and if they work together, two houses take 6 weeks, 6 days to build. Therefore, there is a 1/3 chance that both houses are completed in less than 7 weeks, which corresponds to choice (B).
January 24, 2011
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 number of flips (5) as our n value and the number of heads we want (1) as our k value. We do this because we want to select one flip to be heads. When we plug these numbers into the combinations formula we end up with a result of five. Listed out, these outcomes are HTTTT, THTTT, TTHTT, TTTHT and TTTTH (where T is Tails and H is Heads). The five outcomes that give us one head plus the one outcome that gives us zero heads make six total desired outcomes. But remember, these are the outcomes we do NOT want.
In our coin toss scenario, the total possible outcomes (denominator for our overall probability formula) can be calculated by multiplying the possible outcomes for each individual flip together. Since we have five flips and each flip has two possible outcomes, we have 2x2x2x2x2=32 possible outcomes. If six of these are outcomes we do not want, then 26 are desired. This makes the probability of getting at least two heads 26/32, which can be simplified to 13/16.
If you follow the rules in this article, along with those in the previous articles in this series, you should have the foundation you will need to succeed on GMAT probability problems.
January 19, 2011
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. Another arrangement is A, then B, then D and then C. Another is A, then C, then B and then D and so on. Rather than listing all of these out, we can calculate using factorials. Notated as “n!”, 4! would equal 4 x 3 x 2 x 1 = 24 possible arrangements.
In the second type of question you are not arranging at all, but rather are creating groups. This is a combinations question. Now you could have six people and want to know how many groups of four you could make. If our people are A, B, C, D, E and F, one group would be A, B, C and D. Another group would be A, B, D and F and so on. To calculate how many of these groups you could make use the combinations formula, which is n!/[k!(n-k)!], where n is the number of items with which we are starting (6 in our example) and k is the number of items we want in each group (4 in our example.)
The final scenario you may encounter occurs when you need to both group and arrange. For example, you could have a question that tells you that six people are in a race and the top three are ranked first, second and third and asks how many sets of rankings are possible. Now we need to select three of our six to be ranked and then put them in an order. If A is first, B is second and C is third, then we have a different outcome than if C is first, B is second and A is third. The formula here is n!/(n-k)!, where n and k have the same meanings as above. As long as you are arranging, this is still a type of permutation question.
Thus, when we encounter a combinations and permutations problem we must first identify if we are arranging, grouping or doing both and then apply the correct formula. Next time we will return to probability to see how these skills can be applied there.
January 17, 2011
Now that we have looked at how to handle probability problems that deal with independent events, we are going to consider how the GMAT can make these problems more complex.
The main way in which the GMAT will do this, is by asking you the probability of getting “at least” a certain number of a particular outcome rather than the probability of getting “exactly” that number.
For example, in our last article we considered how to handle a question that asks for the probability of getting exactly one head when flipping a coin twice. A more complex problem would ask us the probability of getting at least two heads when flipping a coin five times. Now, instead of having one desired outcome – exactly one head – we have four outcomes that are considered desired – two, three, four or five heads. To put that in probability speak, we want two heads OR three heads OR four heads OR five heads. As we remember, all of those “ors” become addition.
This is where an important strategy for advanced probability comes into play. In any situation, the sum of the probabilities of all possible events will add up to one. This means that on some GMAT problems, such as the one described above, it will be quicker to find the probability of an event NOT happening and subtracting the result from one, than it would be to find the answer directly. (You can also think of this as “1 minus the probability of something NOT occuring = probability that it WILL occur”.)
The outcomes that do NOT give us one at “least two heads” are: zero heads and one head. We next need to find the probability of each of these events occurring, add those numbers together and subtract from one. But to find those probabilities, we need to veer into the realm of combinations, which will be the topic of the next article in this series. So hold that thought for this question—and stay tuned for our next article!
January 12, 2011
In my last posting, we discussed the basics of probability, which you will build on and use in your GMAT prep. Now, we are going to take those basics one-step further, into the realm of multiple, independent events.
The key to understanding how to deal with multiple events, is to remember these rules: in the world of probability,
- “and” means multiply, and
- “or” means add.
Let’s say you are flipping two coins and you want to know the probability of both of them landing heads up. In probability-speak, you would want to think of this as, “what is the probability of flipping a head on one coin AND a head on the other coin?” In order to determine your answer, you should take the probability of each independent event happening – 1/2 for both flips in this case – and multiply them together. This gives us 1/2 X 1/2 = 1/4.
This gets more complicated when “or” situations come into the picture. Now let’s say we are flipping the same two coins and we want the probability of exactly one of them landing heads up. In this case, we have two different outcomes that give us what we want. We can have either heads on our first flip and tails on our second OR we can have tails on our first flip and heads on our second. This is an “or” situation, which tells us we need to add. But, we also need to account for the “ands” in our sentence. Look at it again. We want heads AND tails OR tails and heads. This means we need to first multiply – remember multiplication comes before addition in the order of operations – and then add. This gives us (1/2 X 1/2) + (1/2 X 1/2) = 1/4 + 1/4 = 1/2, which is the correct answer for this question.
Another way to think about the previous example, is to ask yourself, how many different scenarios would have my desired outcome (exactly 1 head) if I flipped a coin twice? There would be two scenarios of exactly one head (either a head on the first toss, or the second). So 2 becomes the “desired” numerator in the probability ratio. Now, how many “total possible outcomes” are there if we toss a coin twice? Well, you could roll a head then a tail; two heads; two tails; or a tail then a head…so there are 4 distinct possibilities, giving us 2/4, or 1/2, for our final answer.
Note that if you are ever faced with a similar question where there are a lot of possible scenarios, and you don’t want to think through them all or count them all, a shortcut to getting your denominator (total possible outcomes) is to take the number of possible outcomes in any given event (such as 2 outcomes for tossing a coin), raise that to the power of how many times that event occurs (in our example, that is 2 again, since we tossed the coin twice.) So here, there were 2^2 or 4 possible outcomes. If you had tossed the coin 5 times, there would have been 2^5, or 32, possible outcomes, and so on.
Now that we have covered multiple events, in the next article in this series we will look at how to handle more complex probability problems.