Pascal's Triangle in Probability

Watch the following video to learn how Pascal's Triangle Applies to Probability:

So what kind of probability problems can Pascal's Triangle be used?

Pascal's triangle is made up of the coefficients of the Binomial Theorem which we learned that the sum of a row n is equal to 2n. So any probability problem that has two equally possible outcomes can be solved using Pascal's Triangle. In the Problem of Points game explained in the video, the possible outcomes were either heads or tails which both have a probability of .5. Thus, the total amount of different outcomes that could happen with a certain amount of coin flips is 2n which we learned is equal to the sum of the coefficients in the nth row of Pascal's Triangle.

Let's consider the problem where we want to know the probability of flipping exactly 2 heads for 4 coin flips. We know how many total possibilities there are for 4 coin flips by looking at the 4th row (when considering the top row the zero row).

Thus, there are 1+4+6+4+1 = 24 = 16 possible different combinations of heads and tails that can happen when flipping a coin 4 times. To then know how many possible ways there are to get exactly 2 heads, we know that the coefficients of Pascal's triangle are all the coefficients of the Binomial Theorem which are combinations. Since we have four coin flips and are wanting to flip two heads, the number of possible ways we can flip exactly 2 heads is 4C2=6, which is the third term of the 4th row of Pascal's Triangle. There is then a 6/16 chance of flipping exactly two heads when flipping a coin 4 times. If this problem had been at least two heads, you would then add all the terms of Pascal's Triangle that came before the 6. In this case those terms are 4C0=1 which means there are 0 tails and 4 heads and 4C1=4 which is when there is 1 tail and 3 heads flipped. This would then give 11/16.

So, if you wanted to generalize this for any problem, you can find what the total number of possible outcomes there are for repeating an event that has two equally likely outcomes n amount of times, by adding the terms of the nth row of Pascal's Triangle. You would then find what the amount of possible outcomes there are for the event you are concerned with by looking at the term(s) of Pascal's Triangle that associates with the number that you are choosing from the total. In the case with heads or tails, it is the number of heads you are wanting to choose from n amount of flips.


Pascal's Triangle can also be used to solve counting problems where order doesn't matter, which are combinations. It is pretty easy to understand why Pascal's Triangle is applicable to combinations because of the Binomial Theorem. Here are some examples of how Pascal's Triangle can be used to solve combination problems:

Example 1:
There is a group of 10 people that will be picked from to create a committee of 4 people. How many possible different committees of size 4 could be created from 10 people?

In this problem, it does not matter in what order there people are selected, only that there are 4 people selected. So, if the same four people are selected twice this is considered the same combination. Thus, there are 10C4 possible committees. We can then look at the 10th row of Pascal's Triangle and then go over to the 5th term (since the first term is 10C0) and that will give us the number of possible different committees.

Thus, there are 210 possible committees of size 4 that can be created from a selection of 10 people.

Example 2:
Madi and Hannah bought a pack of colorful pens that has 8 pens. They plan on splitting the pack by each getting 4 pens. How many different combinations of size 4 are there that Madi could end up with?

Since they are picking from pens with no assignment or order, this problem can be solved with combinations and thus Pascal's Triangle. Because there are 8 pens and each person chooses 4 of them, the amount of different color combinations are 8C4. To solve this using Pascal's Triangle you would look at the 8th row and then the 5th term.