# Fun With Numbers: Binomial Theorem

I have been quite busy over the last couple of weeks, so I haven’t had too much time to put into my blog. In my little bits of free time, I have put together this short post which talks about a clever trick in mathematics that I have always wanted to share with people. One of the best things about this trick is that it is simple and almost anyone can do it once they learn how. If you are ever around a group of friends and you want to blow them away with your computational speed, this is just the trick* to do that. Before I say more, I first would like you to answer the question below;

The question to the poll below is: What is the coefficient in front of   in the expansion of

Now there are probably only two ways that you answered this question. The first way I would suspect is that you guessed, and the second way would be that you took the long, strenuous, and tedious time of actually working this problem out. It turns out that there is a much quicker way to solve this question.

There is an equation used in probability when dealing with combinations known as “n choose k”. It looks like the following,

Now, thus far in my studies of mathematics, this equation has only appeared a couple of times outside of a probability course. When it does come up, however, it is followed shortly by a great tool in probabilistic mathematics which is the Binomial theorem. The Binomial theorem reads as follows,

We can see pretty easily that this is true. Using a simple example, we see that,

And this is satisfied using the Binomial theorem,

So what if someone asks, what is the coefficient in front of the xy in the expansion of (x+y)^2? Well since is raised to the first power, then our value for is one. Thus, the solution is,

So again I ask you, What is the coefficient in front of   in the expansion of  ?

* I know I call this a trick in this post but it should be noted that this equation was developed with much more serious intent. I use this equation daily by means of solving probability questions and its uses extend beyond into all fields of mathematics.

# Infinite Primes

I’m currently taking a class called computational number theory. In the class we are learning a variety of things from cryptography, quadratic reciprocity, to (the simple sounding) primes. In class we learned about something called a twin prime pair which is a pair of integers p, p+2 which are both prime. Some examples of these are; (3,5), (5,7), (11,13), (17,19), (29,31) …

Now it has been proven that there are infinitely many primes. I will not do the proof unless it is requested, and it is pretty easy to believe given that we have infinite numbers. But are there infinitely many twin primes? Again, simply using our intuitions of probability we want to say yes (given, again, that we have infinite numbers). Amazingly enough however, intuition is not a proof and NOBODY has proved it!

Recently in the world of mathematics however, a man by the name of Yitang Zhang from the University of New Hampshire proved a theorem this year which is similar to the unproven theorem above. What he proved is called the bounded prime gaps conjecture, and what he showed was that there are infinitely many prime numbers p, q where p < q and q – p is less than or equal to 70,000,000.

This is an amazing feat and the proof is said to be upwards of 160 pages long. This also sheds light on some hope that perhaps our tantalizing question above might one day be proven.

On this topic of primes, I will leave you with a fun video. I hope you enjoy.

# Conditional Probability: Is The Fear of Flying Really Irrational?

One of the most common facts said to console someone who fears flying is; “you are more likely be in a car accident on the way to the airport than be in a plane crash.” This is obviously true. The odds of being in a car accident is roughly 1/84 while the odds of being in a plane crash is in the ballpark of 1/500,000.

My mom has a relatively significant fear of flying. The times which she was forced to fly, I remember her trying to fall asleep before lift off while today she tries her hardest to avoid ever flying– I would say she hasn’t flown in the last 15 years or so. Well one day I was pondering to myself about this irrational fear (shortly before I was to be flying to Atlanta) when a thought arose. I wondered; maybe their fears aren’t completely irrational. Maybe the question that is being asked isn’t properly answering what they really fear. Clearly you are more likely to be in a car accident than a plane accident. But you are much more likely to die if you are in a plane crash rather than if you are in a car crash– By roughly 66 times. I think that what aerophobics fear most is dying by plane crash. So I decided to see if this question had any weight to it. Using my mathematical background, I knew which mathematical model could answer this question. The model is known as conditional probability.

Conditional Probability

When first taking a course on probability, conditional probability is one of the first things one begins to learn. It is actually very simple to grasp at first, though the questions can become relatively complex and most universities are aware of all of the harder questions. Lucky for us, the question we are asking today is pretty easy. First however, I need to give a quick introduction to probability.

When determining the probability of an event, questions can be quite elementary. For instance, suppose you have a bag with 4 green balls and 12 red balls. If you put your hand in the bag (and you can’t see inside) and choose a ball at random, what is the probability that you choose a green ball? The answer is simple, there are 16 balls in the bag, 4 of which are green. Therefore the probability you choose a green ball is 4/16 which reduces to 1/4. Conditional probability is just a little more complex.

The easiest way to look at conditional probability is to understand that there are just a few more variables. One of the most basic ways that conditional probability can be stated is this; given two events, A and B, with the probability of B not equal to zero. What is the conditional probability of A, given B? This may sound confusing but I believe an example will help.

EXAMPLE:

Suppose you have 2 bags. Suppose bag 1 contains 4 red balls and 2 green balls, while bag 2 contains 3 red balls and 5 green balls. Suppose you are given a bag and you randomly select a ball out of the bag and it is a red ball. The question can then be asked, given that you selected a red ball, what is the probability it was from bag 2?

Now we need to know some terminology.

P(R) means: Probability of selecting a red ball. Similarly, P(G) would be “math” for “the probability of selecting a green ball”. Now the next one may be difficult if you are not mathematically savvy.

To “say” the question stated above which was; given that you selected a red ball, what is the probability you selected it from bag 2? We would write in “math” language, P(B2|R).

You can almost read that across. what is the (P)robability that you selected from B2 (bag 2) given (think of the vertical line as the word, given, here) that you selected a (R)ed ball.

Now I won’t go into why this is so since this is just a blog post and not a class, but there is a given formula that will answer this question for us which is,

Looking at this equation, I know which values that I need to determine. I need to find the probability of picking a red ball given that I have selected bag 2, I need the probability of selecting bag 2, and then I need to do the same thing, but this time, for bag 1. The solutions respectively are,

Now we Just need to plug these values into the equation above and we get,

And there we have it! We have solved a conditional probability question. Now onto the question that matters.

For more on conditional probability, Wikipedia should suffice.

The Real Question

Like I said at the beginning of this post, it’s about asking the right question. I think what those who fear flying the most are truly afraid of is death. So, to ask the question in terms of conditional probability, the question we are answering is;

Given that you die, what is the probability that you were in a plane crash versus a car crash. Written out mathematically, the question appears as follows,

Let F = fatal, A = car crash, B = plane crash. Then,

vs.

is “math” language for the question above.

We know how to answer this. We can simply follow the example above in order to solve. Now the numbers I found were from reputable sources, but if they were incorrect, that would change the results. For simplicity, we see that answering one of these questions will solve the other. So we can re-ask the question as;

Given that you die, what is the probability it was in a plane crash?

We know from our example above that,

Answering each piece of the formula in order, again by research, I found that the probability of dying given that you are in a plane crash is about 33%, the probability of being in a plane crash is 1/500,000. The probability of dying given that you are in a car accident is about 0.62%, and the probability of being in a car accident is about 1/84. Using these values, we find that…

(Drum roll)

The probability of being in a plane crash, rather than a car crash, given that you died, is…

(Drum roll)

0.9%!!!!!!!

This means that if you take 111 people from a pool of people that died in either a car accident or a plane accident, 110 of them will have died in a car crash. So yes, the answer is very underwhelming, but this should be just another fact that lets all those who have a fear of flying take a breath and relax while looking at the numbers that reassure us even more to the safeties of flying.

A quick story to share with you now that I have finished. I worked on this about a month or two ago and had found a couple different statistics, which I later discovered to be incorrect. But before I knew this, I was given chilling results that suggested the probability in dying in a plane crash versus dying in a car crash was 47.2%. Much more concerning than the unintimidating 0.9%. So lastly, breathe easy, relax, and enjoy your safe flight.