August 20, 2012
Take a look at the picture with this blog. It’s an iconic optical illusion. Stare at it—what do you see? The picture is called the Great Wave off Kanagawa, painted by Katsushika Hokusai, a Japanese artist famed for his brilliant compositions. This drawing is of a wave, of course, but do you see the other wave, the reverse wave in the sky?
This image utilizes negative space. You take the whole frame, the great big rectangle, you block out that actual image—and what remains is, in its own right, an interesting picture.
To find the area of the shaded region, we need to subtract the area of the smaller inner circle from the large outer circle—the difference is the area of the ring.
But the concept extends beyond simple pictures and geometry questions. Probability problems sometimes operate on a similar principle, subtracting an easy-to-find probability of failure from 1, the total of all probabilities, the “whole frame,” as it were. Once you’ve subtracted all the failures, then whatever remains, the “negative space,” must be the chances of success!
The GMAT is a test of critical thinking. It tests your ability to find the most effective path to the solution. Sometimes, you’ll pick numbers, sometimes you’ll do the math directly, sometimes you’ll guess strategically. And sometimes, you’ll define the negative space around the answer, and solve that way. Today’s problem of the day is best solved by that principle. You can’t figure out how many arrangements follow the rule in the question stem. But you can find out how many arrangements don’t follow the rule, and subtract it from the total. Good luck!
Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in
a single row of six chairs. If Betsy cannot sit next to Emily, how many different
arrangements of the six children are possible?
Step 1: Analyze the Question
We have to arrange six children in six chairs, but two of
the children can’t sit together. We’re asked to calculate the
number of different arrangements of children.
Step 2: State the Task
We’ll calculate the number of possible arrangements of the
children. Then, we’ll subtract the number of ways Betsy can
sit next to Emily.
Step 3: Approach Strategically
The possible number of arrangements of six elements is
6! = 6 x 5 x 4 x 3 x 2 x 1= 720.
Now we’ll have to calculate the number of ways that would
violate the question stem by putting Betsy next to Emily.
If we number our seats from left to right, there are 5 ways
they can sit together if Betsy is on the left and Emily is on
Seats 1 & 2
Seats 2 & 3
Seats 3 & 4
Seats 4 & 5
Seats 5 & 6
And there are 5 more ways if Emily is on the left and Betsy
is on the right, for a total of 10. Now, for any one of those
10 ways, the four remaining children can be seated in 4!
ways: 4! = 4 x 3 x 2 x 1 = 24. So we need to subtract
24 x 10 = 240 ways that have Betsy and Emily sitting
together from our original total of 720: 720 – 240 = 480.
Answer choice (B).
August 16, 2012
Tackling some of the tougher GMAT probability questions efficiently relies on both steady practice and your ability to make two key decisions well. First, you will need to quickly and accurately assess the total number of possible outcomes (the denominator of your probability equation). Second, within a multitude of possible approaches, you will need to determine the most efficient route to calculate the number of desired outcomes (the numerator of your probability equation).
With the clock ticking away on your GMAT CAT, figuring out the total number of possibilities can be time-consuming and fraught with room for error. For instance, if a question asks about the probability of getting at least 2 heads on 5 coin tosses, you could sit there all day writing out possibilities:
So forth and so on. I know I got dizzy with the possibilities just writing those three out. There is a better and more efficient way. For every coin that you toss there are 2 possibilities. You can think of the total possibilities like a permutation problem.
__2__ __2__ __2__ __2__ __2_
1st 2nd 3rd 4th 5th
Just like in a GMAT permutations question when we are trying to determine the total number of codes possible or 4-digit numbers, we would multiply these individual probabilities together. Therefore, there are 2x2x2x2x2 = 2^5 = 32 total possibilities.
Next, we need to look at the numerator (desired outcomes). We want to find the all of the possibilities that have at least 2 heads, which means that we could have 2 heads, 3 heads, 4 heads, or 5 heads. To do so, we would need to count all of the different ways that these possibilities could be arranged. Again, we find ourselves in a situation that will be time-consuming and fraught with error. Instead of going down this path, remember that the sum of the probabilities of a complete set of mutually exclusive possible outcomes is 1. Thus, as is often the case on “at least” probability questions, we can look for those options that are restricted. Then we only have to count the options that have 1 or 0 heads.
There are only 6 of those, instead of the 26 possibilities the other way.
Finally, we can either subtract 6/32 from 1 in order to remove all of the restricted possibilities from 1 or we can subtract 6 from 32 and use the result as the desired possibilities. Either way, the answer is 26/32, which you can reduce down to 13/16.
Let’s look at another to make sure we have this down.
A test has 5 multiple-choice questions. Each question has 4 answer options (A,B,C,D). What is the probability that a student will choose “B” for at least four questions if she leaves no questions blank? Pause a moment and try it for yourself first.
Step 1: Total number of possibilities
There are 5 questions and each has 4 possibilities, so our total possibilities would be 4x4x4x4x4 = 4^5 = 1024
Step 2: Approach Desired Strategically
Here there are far more possibilities for 0, 1, 2, or 3 “B’s,” so let’s get a total for 4 or 5 “B’s”.
All B’s — B,B,B,B,B
Four B’s — A,B,B,B,B – B,A,B,B,B – B,B,A,B,B – B,B,B,A,B — B,B,B,B,A
C,B,B,B,B – B,C,B,B,B – B,B,C,B,B – B,B,B,C,B – B,B,B,B,C
D,B,B,B,B – B,D,B,B,B – B,B,D,B,B – B,B,B,D,B – B,B,B,B,D
3 x 5 = 15 because we can repeat the same pattern for each letter other than B
We can also calculate the total possibilities of 4 B’s by calculating the possibilities for each “no-B” position.
No-B,B, B, B, B = 3x1x1x1x1 = 3
B, No-B, B, B, B = 1x3x1x1x1 = 3
B, B, No-B, B, B = 1x1x3x1x1 = 3
B, B, B, No-B, B = 1x1x1x3x1 = 3
B, B, B, B, No-B = 1x1x1x1x3 =3
A total of 15 possibilities with 4 B’s in the mix
That gives a total of 16 different ways that a student can choose at least 4 B’s here.
16/1024 = 1/64 as our final probability.
Keep these two decisions in mind each time that you approach a tough probability question on the GMAT quantitive section. You don’t have to write out all of the possible outcomes in order to tackle these on test day!
The language of probability can take a while to learn, especially if you are unfamiliar or out of practice with it to start. Post your questions below, and we can help you get on track.
August 8, 2012
Translating word problems into algebra is a staple skill of GMAT test-takers, one that underlies countless problems in practice and on Test Day. But some challenging translations occur as part of probability and combinatorics problems. That’s because a pair of the most basic words in the English language, “And” and “Or,” suddenly become overburdened with mathematical significance.
“And” is the simpler of the two. When “And” represents independent choices—cases in which one option or arrangement has no impact on the other choice—just multiply the outcomes. For instance:
“The number of ways to purchase three board games and two video games” is an independent choice. The board games we pick have no impact on the video games we pick. So, to translate: [The number of ways to purchase three board games] × [the number of ways to select two video games]. Of course, we’d need the combination formula to find actual values—but we’d know what to do with those values once we got them.
“Or” is a little more complicated. It’s confusing even in conversation, after all—if I say that you can have cake or ice cream for dessert, can you have both if you want? When you CAN have both, you can treat the problem similarly to an overlapping sets problem. But in most cases on the GMAT, the “Or”s will be mutually exclusive—for instance, if you want to know the odds of drawing a heart or a diamond out of a deck of cards, there is no card that is both a heart or diamond.
A mutually exclusive OR can be translated as a “plus.” That’s all you have to do. So:
“The probability of drawing a heart or a diamond from a deck of cards,” which is the odds of one of two mutually exclusive events occurring, translates to: [The probability of drawing a heart] + [The probability of drawing a diamond].
Today’s problem of the day hinges on those same ideas. Read carefully—you’re solving for the odds of one of two outcomes (an OR), but each of those two outcomes is the specific result of two independent events (an AND). Be systematic in your translation, and I’m sure you’ll get the right result.
Post your answers below before you read the solution, and we can go over them…
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?
Step 1: Analyze the Question
This is a complex question, but it can be broken down into
simple steps. As with any probability question, we must first
consider all of the scenarios in which the desired outcome
can be true. In this question, there are two different ways
in which exactly one of two students chosen is a senior.
Either (i) a senior is chosen from Room A and a non-senior
is chosen from Room B or (ii) a non-senior is chosen from
Room A and a senior is chosen from Room B.
Step 2: State the Task
Determine the probabilities of the two scenarios above and
add them together.
Step 3: Approach Strategically
Let’s start with (i) and find the probability that a senior is
chosen from Room A and a nonsenior is chosen from Room B.
The probability that the student chosen from Room A is a
senior is 1/6 .
The probability that the student chosen from Room B is not
a senior is 1- 5/7=2/7
So the probability that the student chosen from Room A
is a senior and the student chosen from Room B is not a
senior is (1/6) x (2/7) = 2/42 .
Let’s not simplify this yet, because we can expect that the
probability we will find when working with (ii) will also
have a denominator of 42.
Now let’s work with (ii). Let’s find the probability that a
nonsenior is chosen from Room A and a senior is chosen
from Room B.
The probability that the student chosen from Room A is not
a senior is 1 – 1/6 = 5/6 .
The probability that the student chosen from Room B is a
senior is 5/7 .
So the probability that the student chosen from Room A
is a not a senior and the student chosen from Room B is a
senior is (5/6) x (5/7) = 25/42 .
Now we sum the total desired outcomes. The probability
that exactly one of the students chosen is a senior
is (2/42) + (25/42) = 27/42 = 9/14 .
(C) is correct.
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).