Augie’s Socks: Part II


In part one, I introduced this set of facts:

Augie is blind. Thanks to his organized systems he is able to live alone. However, one day he finds that his laundry service has forgotten to pair his socks together. Augie has 12 black socks, 8 white socks, 6 blue socks and 4 red socks. Augie puts two random socks on.

Now, let’s ask a probability question.

What is the probability that Augie has put on a matching pair of socks?

This question tests both OR and AND probabilities. Let’s lay it out systematically.

There are four ways Augie could be wearing a matching pair of socks (black, black), (white, white), (blue, blue) or (red,red).

Since any one of these situations satisfies our condition of matching socks we want to add their individual probabilities together.

P(matching black) + P(matching white) + P(matching blue) + P(matching red) = P(matching pair of socks)

Now, we need to find those four probabilities individually. In order to pull a matching pair of black socks. In order to do that the first sock we pull out must be black. The odds of that are 12/30 (number of black socks/number of total socks) (12/30 reduces to 2/5 but we’ll save reducing for the end). But in order to pull a pair of black socks we need to pull a black sock first AND pull a black sock second. In order to pull the second black sock we’ll have to pull one of the remaining 11 black socks out of the remaining 29. Since we need both events to occur in order to get a black pair, we’ll multiply the individual probabilities to get the probability of getting a black pair:

12/30 * 11/29 = 132/870

Following the same process for the other colors gets you:

P(matching white) = 8/30 * 7/29 = 56/870

P(matching blue) = 6/30 * 5/29 = 30/870

P(matching red) = 4/30 * 3/29 = 12/870

Adding the probabilities gets a total of 230/870 = 23/87 or about 26.4%.

In this case you need to remember that when there are two individual events that must BOTH occur, you multiply their probabilities together. When there are multiple desired outcomes that are EACH satisfactory you add their probabilities.

 Augie’s Socks: Part I


Augie is blind. Thanks to his organized systems he is able to live alone. However, one day he finds that his laundry service has forgotten to pair his socks together. Augie has 12 black socks, 8 white socks, 6 blue socks and 4 red socks. Augie puts two random socks on and decides to take several more to work with him in his briefcase so that he can change once he gets to work and consults with his co-worker Annie if he has picked a pair that doesn’t match. How many socks must Augie put in his briefcase to ensure that he won’t go through the day with mis-matched socks?

A) 1

B) 2

C) 3

D) 5

E) 25

There are two levels to this problem. The first is recognizing this is a worst-case scenario problem. It’s more logic than math. When asked to consider how many socks you need to ensure Augie has a pair of matching socks, you should think about the maximum number of socks Augie could have without making a match. In this case, there are four different colored socks so Augie could end up with four socks and still not have a match. However, if we pull a fifth sock, no matter what color it is, that sock will match one that we already have in the worst case scenario.

That’s the first level, and students who stop there usually end up with answer choice D. The second level requires careful reading as you notice that the question asks how many socks Augie needs to put in his briefcase. Remember, he already has two socks on his feet! To get to the total of five socks that we need, we only need 3 socks in the briefcase.

Remember, the correct answers to math questions often turn on logic and careful reading rather than math skills!

In part two, I’ll show how a similar fact pattern can be used in a probability problem.

 10 GRE Resolutions for the New Year


1. I resolve to make sleep a priority. Sure, sometimes right before bed is the only time I can study, but when I jeopardize my sleep I make everything more difficult for myself, life gets increasingly stressful and I end up studying even less.

2. I resolve to read the instructions (once). I’m pretty sure I know how to do these problems, but I’m going to sit down one time now and read the instructions for every section so that I sure I really understand what the test is asking me to do. I won’t risk letting my assumptions get me in trouble.

3. I resolve to become comfortable with questions will multiple answers. Although the GRE is throwing something a little different at me, I will take the time to refine my strategies for dealing with these questions.

4. I resolve to focus on the goal. As important as GRE scores are to me, I will view them as a means to an end. Admission to my preferred program will be my focus, not some artificially high GRE score that I feel like I need.

5. I resolve to study better, not more. Rather than spending three hours with my books open while Real Housewives is on in the background, I will do my studying in manageable chunks of time without anything else to distract me.

6. I resolve to read more. I understand that reading comprehension is a skill that is best developed by reading. I will check out my local library, find some good books and enhance the mental pathways that lead to better scores.

7. I resolve to set a test date and stick with it. I know that as my test date approaches I will get nervous and worry that I’m not ready. I also know that if I view my test date as flexible I won’t devote myself to being prepared as well as I should.

8. I resolve to create a list of unfamiliar words that I come across. I know the GRE tests a lot of unfamiliar vocabulary, but it tends to test the same set of words. By being intentional about cataloging the words that I come across, I will create a valuable study tool.

9. I resolve to stay positive. Even when practice scores aren’t as high as I want them to be. Even when I’m tired and frustrated.

10. I resolve to keep balance in my life. The GRE is an important part of my life, but it is not my whole life. In this coming year I will keep a good balance between study, work, school, friends, family and fun.

Happy 2015!

 System of Equations Problem


What is the solution to the system of equations shown below?

6x – 3y = -12

4x + 3y = 2

A) (2,2)

B) (-2,-1)

C) (-1,-1)

D) (-1,2)

E) (1,-2)

In order to solve this problem, we first need to understand what it’s asking. So, what is a “system of equations”? Well, that simply means that we have more than one equation that deals with the same variables. Don’t let that name overwhelm you into thinking it’s more complicated than it really is! In this case, we have to linear equations and solving a system of linear equations simply means finding the point where the two lines intersect. At that point you’ll get an x-value and a y-value that work in both equations.

There are several different ways to solve a system of linear equations and trick is to figure out which one is most efficient for the equations that you’re given. The three methods are:

1. Graphing- If you turn your equations into y=mx+b form you should be able to graph them out and figure out where they intercept. The advantage of this method is that it’s very straight-forward. The disadvantage is that it requires you to be very precise with the lines you draw and can be infeasible when you have an intersection that’s relatively far from the origin. Graphing is typically best used to estimate, or verify a solution.

2. Substitution- This method helps you turn unruly two-variable equations into single-variable equations that can easily be solved. You do this by finding one variable in terms of the other. Let’s see how that works:

Take the first equation.

6x – 3y = -12

Solve for one of the variables.

3y = 6x + 12

y= 2x +4

Then, plug that value into the second equation.

4x + 3(2x+4) = 2

4x+ 6x+ 12 = 2

10x = -10

x = -1

Now that you have the value for x, you can plug that back into either one of your equations to find the value for y.

4(-1) +3y = 2

3y= 6

y= 2

The great part about substitution is that you know it’s going to work. The downside is that there are quite a few steps in the process so it can be somewhat time consuming. That’s where our third method comes in.

3. Combination- This method involves adding or subtracting the two equations with each other with the goal of eliminating one of the variables. When combination can be done without too much manipulation, it’s the easiest and quickest way to solve a system of equations. That’s the case here.

6x – 3y = -12

+ 4x + 3y = 2


10x = -10

As you can see here, when we add the two equations together the y variable drops out and you have the value for x so you can solve from there.

Remember not to worry when you see a system of equations. Choose the best method to solve based on the problem you’re given and you’re all set!