Now and Venn


One of the books I’m currently reading is Priceless: The Myth of Fair Value (and How to Take Advantage of It) by William Poundstone. It’s an interesting look as some of the psychology behind how we think about prices and perceive value. Poundstone recounts one experiment done by Kahneman and Tversky called “Linda the Bank Teller” that I’d like you to think about.

“Linda is 31 years old, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.”

Which of the following is more likely to be true?

Linda is a bank teller.

Linda is a bank teller and is active in the feminist movement.

Astonishingly, 85% of the college students tested responded that the second statement is more likely to be true than the first, which is absolutely ridiculous. The point that Kahneman and Tversky were making is that we rely on heuristics or “rules of thumb” more often than is logically sound. The point I’d like to make is about drawing diagrams.

The information you’re given in this situation, although different in format from what you’re going to see on a standardized test is nonetheless similar. You’re given a few different pieces of information that fit together in some way that may not be intuitively obvious. One of the best ways to deal with that situation is to draw a graphical representation of what you’ve been told. When given two overlapping sets, you should draw a Venn diagram.

Now, let’s consider the two situations. First, the chance that Linda is a bank teller is represented by the region in red.

The chance that Linda is a bank teller and is active in the feminist movement is represented by the region in blue.

As you can see, the second set is a subset of the first. A drawing with only the two regions we’re concerned with would look something like this:

Linda can’t be a feminist and bank teller without being a bank teller. In this graphic form it’s pretty obvious, and yet 85% of the students missed this fact the first time around. This is a great example of the benefits of converting words into graphical forms so that you can make sure you have a good understanding of what you’re doing before you go forward with a problem.

 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.

 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!