An “Undesired” Approach to GMAT Probability gmat article
Posted on
11
Mar 2021

An “Undesired” Approach to GMAT Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 11th March, 2021

In our last post, we discussed a solution for the following question, which is a twist on an Official Guide GMAT probability problem:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

We also mentioned that there was an alternate way to solve it. Did you find it? In truth, it relates to something we discussed in a previous post we did on GMAT Combinatorics, specifically Combinations with Restrictions. In that post, we discussed the idea of considering combinations in which you’re not interested. It might seem counterintuitive, but if you subtract those out from the total number of combinations possible, you’re left with the number of combinations in which you are interested:

You can actually do something similar with probability. Take the following basic example:

Suppose I told you to flip a fair coin five times, “fair” meaning that it has an equal chance of landing heads-up or tails-up. I then wanted to know the probability that I flip at least one head. Now, when you think about it, the language “at least one” involves so many desired possibilities here. It could be 1 head, 2 heads, …, all the way up to 5 heads. I’d have to calculate each of those probabilities individually and add them up.

Or…

I could consider what is not desired, since the possibilities are so much fewer:

0 heads   |   1 head      2 heads      3 heads      4 heads      5 heads

All of the above must add to 100% or 1, meaning all possible outcomes. So why not figure out the probability that I get 0 heads (or all tails), and then subtract it from 100% or 1 (depending on whether I’m using a percentage or decimal/fraction)? I’ll then be left with all the possibilities in which I’m actually interested, without the need to do any more calculations.

Each time I flip the coin, there is a ½ chance that I flip a tail. This is the same each of the five times I flip the coin. I then multiply all of the probabilities together:

½ x ½ x ½ x ½ x ½ = 1 / 25  = 1 / 32

Another way to view this is through combinatorics. Remember, probability is always Desired outcomes / Total possible outcomes. If we start with the denominator, there are two outcomes each time we flip the coin. That means for five flips, we have 25 or 32 possible outcomes, as illustrated here with our slot method:

_2_  _2_  _2_  _2_  _2_ = 32

Out of those 32 outcomes, how many involve our (not) desired outcome of all tails? Well, there’s only one possible way to do that: 

_T_  _T_  _T_  _T_  _T_    ← Only 1 outcome possible

It really is that straightforward: with one outcome possible out of 32 total, the probability is 1/32 that we flip all tails. 

Now remember, that is our, not desired. Our desired is the probability of getting at least one head

0 heads   |   1 head      2 heads      3 heads      4 heads      5 heads

So, since the probability of getting 0 heads (all tails) is 1/32, we simply need to subtract that from 1 (or 32/32) to get our final result. The probability that we flip at least one head if we flip a fair coin five times is 31/32.

Application to problem from previous post

So now, how do we work that into the problem we did last time? Well, in the previous post, we took a more straightforward approach in which we considered the outcomes we desired. But can we use the above example and consider not desired instead? Think about it and give it a shot before reading the explanation:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B) 5/8
C) 3/8
D) 5/64
E) 3/64

Explanation

In this question, our desired case is that Zelda does not attempt the problem. That means, quite simply, that our not desired case is that Zelda does get to attempt it. This requires us analytically to consider how such a case would arise. Let’s map out the possibilities with probabilities:

An “Undesired” Approach to GMAT Probability treeNotice that two complementary probabilities are presented for each box. For example, since there is a 1/4 chance Xavier solves the problem (left arrow), we include the 3/4 probability that he does not solve the problem (right arrow). 

If Zelda does get to attempt it, it’s clear from the above that first Xavier and Yvonne must each not solve it. There is a 3/4 and a 1/2 chance, respectively, of that happening. This is also a dependent situation. Xavier must not solve AND Yvonne must not solve. Therefore, we will multiply the two probabilities together to get ¾ x ½ = ⅜. So there is a 3/8 chance of getting our not desired outcome of Zelda attempting the problem.

So, we can finally subtract this number from 1 (or 8/8) and see that there is a 5/8 chance of Zelda not getting to attempt the problem. The correct answer is B.

Next time, we’ll discuss how GMAT Probability and Combinatorics can combine to form some interesting problems…

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

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Independent vs Dependent Probability article for the GMAT
Posted on
09
Mar 2021

Independent vs. Dependent Probability

By: Rich Zwelling, Apex GMAT Instructor
Date: 8th March, 2021

Independent vs. Dependent Probability

As promised last time, we’ll return to some strict GMAT probability today. Specifically, we’ll discuss the difference between independent and dependent probability. This simply refers to whether or not the events involved are dependent on one another. For example, let’s take a look at the following Official Guide problem:

Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1/4, 1/2 and 5/8, respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem?

A) 11/8
B)
7/8
C) 9/64
D) 5/64
E) 3/64

In this case, we are dealing with independent events, because none of the probabilities affect the others. In other words, what Xavier does doesn’t affect Yvonne’s chances. We can treat each of the given probabilities as they are. 

So mathematically, we would multiply, the probabilities involved. (Incidentally, the word “and” is often a good indication that multiplication is involved. We want Xavier AND Yvonne AND not Zelda to solve the problem.) And if Zelda has a chance of solving the problem, that means she has a chance of not solving it. 

The answer would therefore be ¼ x ½ x ⅜  = 3/64 or answer choice E. 

What if, however, we changed the problem to look like this:

Xavier, Yvonne, and Zelda individual probabilities for success on a certain problem are 1/4, 1/2 and 5/8, respectively. Xavier will attempt the problem first. If he solves it, Yvonne and Zelda will not attempt it. If Xavier cannot solve it, Yvonne will attempt it next. If she solves it, Zelda will not attempt it. If Yvonne cannot solve it, Zelda will then attempt it. What is the probability that Zelda does not get to attempt the problem?

A) 3/16
B)
5/8
C) 3/8
D) 5/64
E) 3/64

As you can see, the problem got much more complicated much more quickly, because now, the question stem is dependent upon a very specific series of events. Now, the events affect one another. Xavier will attempt the problem, but what happens at this stage affects what happens next. If he solves it, everything stops. But if he doesn’t, the problem moves to Yvonne. So in effect, there’s a ¼ chance that he’s the only person to attempt the problem, and there’s a ¾ chance the problem moves to Yvonne.

This is most likely how the GMAT will force you to think about probability: not in terms of formulas or complicated mathematical concepts, but rather in terms of narrative within a new problem with straightforward numbers. 

That brings us to consideration of the question stem itself. What would have to happen for Zelda not to attempt the problem? Well, there are a couple of possibilities:

 1. Xavier solves the problem

If Xavier solves the problem, the sequence ends, and Zelda does not see the problem. This is one case we’re interested in, and there’s a ¼ chance of that happening. 

 2. Xavier does not solve, but then Yvonne solves

There’s a ½ chance of Yvonne solving, but her seeing the problem is dependent upon the ¾ chance that Xavier does not solve. So in reality, we must multiply the two numbers together to acknowledge that the situation we want is “Xavier does not solve AND Yvonne does solve.” This results in ¾ x ½ = ⅜ 

The two above cases constitute two independent situations that we now must add together. For Zelda not to see the problem, either Xavier must solve it OR Yvonne must solve it. (The word “or” is often a good indication that addition will be used).

This leads us to our final probability of ¼ + ⅜ = that Zelda does not get to attempt the problem.

There is an alternative way to solve this problem, which we’ll talk about next time. It will segue nicely into the next topic, which we’ve already hinted at in our posts on GMAT combinatorics. Until then…

Permutations and Combinations Intro
A Continuation of Permutation Math
An Intro To Combination Math
Permutations With Repeat Elements
Permutations With Restrictions
Combinations with Restrictions
Independent vs Dependent Probability
GMAT Probability Math – The Undesired Approach
GMAT Probability Meets Combinatorics: One Problem, Two Approaches

Read more
Triangle Inequality Rule on the GMAT
Posted on
09
Feb 2021

Triangle Inequality Rule

By: Rich Zwelling, Apex GMAT Instructor
Date: 9th February, 2021

One of the less-common but still need-to-know rules tested on the GMAT is the “triangle inequality” rule, which allows you to draw conclusions about the length of the third side of a triangle given information about the lengths of the other two sides.

Often times, this rule is presented in two parts, but I find it is easiest to condense it into one, simple part that concerns a sum and a difference. Here’s what I mean, and we’ll use a SCENARIO:

Suppose we have a triangle that has two sides of length 3 and 5:

triangles inequalities 1

What can we say about the length of the third side? Of course, we can’t nail down a single definitive value for that length, but we can actually put a limit on its range. That range is simply the difference and the sum of the lengths of the other two sides, non-inclusive.

So, in this case, since the difference between the lengths of the other two sides is 2, and their sum is 8, we can say for sure that the third side of this triangle must have a length between 2 and 8, non-inclusive. [Algebraically, this reads as (5-3) < x < (5+3) OR 2 < x < 8.]

If you’d like to see that put into words:

**The length of any side of a triangle must be shorter than the sum of the other two side lengths and longer than the difference of the other two side lengths.**

It’s important to note that this works for any triangle. But why did we say non-inclusive? Well, let’s look at what would happen if we included the 8 in the above example. Imagine a “triangle” with lengths 3, 5, and 8. Can you see the problem? (Think about it before reading the next paragraph.)

Imagine a twig of length 3 inches and another of length 5 inches. How would you form a geometric figure of length 8 inches? You’d simply join the two twigs in a straight line to form a longer, single twig of 8 inches. It would be impossible to form a triangle with a side of 8 inches with the original two twigs.

triangle inequalities 2

 

If you wanted to form a triangle with the twigs of 3 and 5, you’d have to “break” the longer twig of 8 inches and bend the two twigs at an angle for an opportunity to have a third side, guaranteed to be shorter than 8 inches:

triangle inequalities 3

The same logic would hold for the other end of the range (we couldn’t have a triangle of 3, 5, and 2, as the only way to form a length of 5 from lengths of 2 and 3 would be to form a longer line segment of 5.)

Now that we’ve covered the basics, let’s dive into a few problems, starting with this Official Guide problem:

If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2, 7, and k?
(A) one
(B) two
(C) three
(D) four
(E) five

Strategy: Eliminate Answers

As usual with the GMAT, it’s one thing to know the rule, but it’s another when you’re presented with a carefully worded question that tests your ability to pay close attention to detail. First, we are told that two of the lengths of the triangle are 2 and 7. What does that mean for the third side, given the triangle inequality rule? We know the third side must have a length between 5 (the difference between the two sides) and 9 (the sum of the two sides).

Here, you can actually use the answer choices to your advantage, at least to eliminate some answers. Notice that k is specified as an integer. How many integers do we know now are possible? Well, if k must be between 5 and 9 (and remember, it’s non-inclusive), the only options possibly available to us are 6, 7, and 8. That means a maximum of three possible values of k, thus eliminating answers D and E.

Since the GMAT is a time-intensive test, you might have to end up guessing now and then, so if you can strategically eliminate answers, it increases your chances of guessing correctly.

Now for this problem, there’s another condition given, namely that 2 < k < 7. We already determined that k must be 6, 7, or 8. However, of those numbers, only 6 fits in the given range 2 < k < 7. This means that 6 is the only legal value that fits for k. The correct answer is A.

Note:

It’s important to emphasize that the eliminate answers strategy is not a mandate. We’re simply presenting it as an option that works here because it is useful on many GMAT problems and should be explored and practiced as often as possible.

Check out the following links for our other articles on triangles and their properties:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Triangles with Other Shapes
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

Read more
similar triangles on the gmat
Posted on
02
Feb 2021

Similar Triangles – GMAT Geometry

By: Rich Zwelling, Apex GMAT Instructor
Date: 2nd February, 2021

One of the most important things to highlight here is that “similar” does not necessarily mean “identical.” Two triangles can be similar without being the same size. For example, take the following:

similar triangles on the GMAT 1

Even though the triangles are of different size, notice that the angles remain the same. This is what really defines the triangles as similar.

Now, what makes this interesting is that the measurements associated with the triangle increase proportionally. For example, if we were to present a triangle with lengths 3, 5, and 7, and we were to then tell you that a similar triangle existed that was twice as large, the corresponding side lengths of that similar triangle would have to be 6, 10, and 14. (This should be no surprise considering our lesson on multiples of Pythagorean triples, such as 3-4-5 leading to 6-8-10, 9-12-15, etc.)

You can also extend this to Perimeter, as perimeter is another one-dimensional measurement. So, if for example we ask:

similar triangles on the GMAT 2

A triangle has line segments XY = 6, YZ = 7, and XZ = 9. If Triangle PQR is similar to Triangle XYZ, and PQ = 18, as shown, then what is the perimeter of Triangle PQR?

Answer: Perimeter is a one-dimensional measurement, just as line segments are. As such, since PQ is three times the length of XY, that means the perimeter of Triangle PQR will be three times the perimeter of Triangle XYZ as well. The perimeter of Triangle XYZ is 6+7+9 = 22. We simply multiply that by 3 to get the perimeter of Triangle PQR, which is 66.

Things can get a little more difficult with area, however, as area is a two-dimensional measurement. If I double the length of each side of a triangle, for example, how does this affect the area? Think about it before reading on…

SCENARIO

Suppose we had a triangle that had a base of 20 and a height of 10:

similar triangles on the GMAT 3

The area would be 20*10 / 2 = 100.

Now, if we double each side of the triangle, what effect does that have on the height? Well, the height is still a one-dimensional measurement (i.e. a line segment), so it also doubles. So the new triangle would have a base of 40 and a height of 20. That would make the area 40*20 / 2 = 400.

Notice that since the original area was 100 and the new area is 400, the area actually quadrupled, even though each side doubled. If the base and height are each multiplied by 2, the area is multiplied by 22. (There’s a connection here to units, since units of area are in square measurements, such as square inches, square meters, or square feet.)

Now, let’s take a look at the following original problem:

Triangle ABC and Triangle DEF are two triangular pens enclosing two separate terrariums. Triangle ABC has side lengths 7 inches, 8 inches, and 10 inches. A beetle is placed along the outer edge of the other terrarium at point D and traverses the entire perimeter once without retracing its path. When finished, it was discovered that the beetle took three times as long as it did traversing the first terrarium traveling at the same average speed in the same manner. What is the total distance, in inches, that the beetle covered between the two terrariums?

A) 25
B) 50
C) 75
D) 100
E) 125

Explanation

This one has a few traps in store. Hopefully you figured out the significance of the beetle taking three times as long to traverse the second terrarium at the same average speed: it’s confirmation that the second terrarium has three times the perimeter of the first. At that point, you can deduce that, since the first terrarium has perimeter 7+8+10 = 25, the second one must have perimeter 25*3 = 75. However, it can be tempting to then choose C, if you don’t read the question closely. Notice the question effectively asks for the perimeters of BOTH terrariums. The correct answer is D.

GMAT Triangle Series Articles:

A Short Meditation on Triangles
The 30-60-90 Right Triangle
The 45-45-90 Right Triangle
The Area of an Equilateral Triangle
Triangles with Other Shapes
Isosceles Triangles and Data Sufficiency
Similar Triangles
3-4-5 Right Triangle
5-12-13 and 7-24-25 Right Triangles

Read more