Lesson 49: Using the Binomial Theorem

New Concept

If $n$ is a nonnegative integer, then:

What’s next

Next, you'll apply this theorem to expand binomials and solve probability problems.

Pascal's Triangle is a triangular array where each number is the sum of the two directly above it. The numbers in the nth row are the coefficients for the terms in the expansion of the binomial $(a+b)^n$. The top row is considered the zero row.

Expand $(x+y)^3$: Use row 3 (1, 3, 3, 1) to get $1x^3 + 3x^2y + 3xy^2 + 1y^3$.
Expand $(a+b)^4$: Use row 4 (1, 4, 6, 4, 1) to get $a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$.

Think of this triangle as a visual shortcut for expanding binomials. Instead of tediously multiplying $(a+b)$ by itself, you just find the correct row for your power and use those numbers as coefficients. It’s a handy map that lays out all the numbers you need to quickly solve the expansion problem.

If $n$ is a nonnegative integer, then $(a+b)^n = ({}_nC_0)a^nb^0 + ({}_nC_1)a^{n-1}b^1 + ... + ({}_nC_n)a^0b^n$. This can be written as $\sum_{r=0}^{n} ({}_nC_r)a^{n-r}b^r$, where ${}_nC_r = \frac{n!}{r!(n-r)!}$.

Expand $(x+2)^3$: This becomes $({}_3C_0)x^3(2)^0 + ({}_3C_1)x^2(2)^1 + ({}_3C_2)x^1(2)^2 + ({}_3C_3)x^0(2)^3 = x^3 + 6x^2 + 12x + 8$.

Expand $(2x-y)^3$: This becomes $({}_3C_0)(2x)^3(-y)^0 + ({}_3C_1)(2x)^2(-y)^1 + ({}_3C_2)(2x)^1(-y)^2 + ({}_3C_3)(2x)^0(-y)^3 = 8x^3 - 12x^2y + 6xy^2 - y^3$.

If $p$ is the probability of success and $q$ is the probability of failure in one trial of a binomial experiment, then the binomial probability of exactly $n$ successes in $m$ trials is given by ${}_mC_n p^n q^{m-n}$.

Probability of 4 heads in 6 coin tosses: $P(\text{4 heads}) = ({}_6C_4)(\frac{1}{2})^4(\frac{1}{2})^2 = 15 \cdot \frac{1}{16} \cdot \frac{1}{4} = \frac{15}{64}$.
Probability of rolling a 6 exactly twice in 5 rolls of a die: $P(\text{2 sixes}) = ({}_5C_2)(\frac{1}{6})^2(\frac{5}{6})^3 = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} \approx 0.16$.

Ever wonder the odds of getting exactly 3 tails in 5 coin flips? This formula is your answer! It's designed for any experiment with exactly two outcomes, like heads or tails. It calculates the precise probability of a specific number of 'successes' happening over a set number of trials, which is super useful in games and stats.