Explanation of the formula for binominal distribution

Question

$$P(Y=k)=^nC_k*p^k*(1-p)^{n-k}$$

$$\\$$

What was the logical thinking behind this formula?

• I know that $^nC_k$ is the total amount of sets
• I know that $p^k*(1-p)^{n-k}$ is the probability of a certain thing happening, and you multiply this by the total amount of sets to get in how many of those sets this thing happens
• I also know that this formula has to do with Pascal's triangle.

How does the logic of Pascal's triangle apply to this formula? How did someone come up with it?

Answer 1

I like this explanation I give. Let $X$ be a random variable that follows the binomial distribution with parameters $n$ and $p$ (number of trials and probability of success, respectfully). Then

$$ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}. $$

Here's each piece:

First, $\binom{n}{x}$ is the number of ways to arrange the successes, in which their ordering does not matter. That's how we count with combinations.

Next, $p^x$ is the probability of all the successes. Finally, $(1-p)^{n-x}$ is the probability of all the failures. And there you go.

Answer 2

We know $ (a + b)(c + d) = ac + ad + bc + bd$, How do you come up with final result. Its just selecting each time $ a\ or\ b $ and then $c\ or\ d$.

For more intuition consider creating a tree with two branches a, b then each of these branch has two branches c and d. Multiply entries from root to leaf and sum them. You got your result.

Consider $(x + y)^n = (x + y) (x + y) ... n\ times$

Let's say $n = 5$ then one of the entry is $(x, x, x, y, y)$. Now for this order doesn't matter. Hence, total occurrences for this $\frac{5!}{3!2!}$ which is ${5 \choose 3}$.

Hope I answer your question.