Sir Francis Galton, 1822-1911, a British mathematician, invented a nifty toy, which demonstrates clearly that a series of pairs of outcomes (ball bouncing to left or right on pins) will produce the pattern known as the binomial distribution.

$$P(X=r)=(\table n;r)p^rq^{n-r}$$where r = 0, 1, 2, ...., n, and $(\table n;r)$ ≡ nCr, $P(X=r) ≡ P_r$, and $p$ and $q$ are the respective probabilities of outcomes event $p$ and event $q$.

n = 0 1 n = 1 1 1 n = 2 1 2 1 n = 3 1 3 3 1 n = 4 1 4 6 4 1 n = 5 1 5 10 10 5 1 .... n = n $({\table n;0})$ $({\table n;1})$ $({\table n;2})$ ... $({\table n;{n-2}})$ $({\table n;{n-1}})$ $({\table n;n})$

These are the binomial coefficients of the expansion of any expression to the power of n.

where $({\table n;r})$ is the binomial coefficient of $a^{n-r}b^r$, and r is any integer from 0 to max. n.

The general term in the binomial expansion of $(a + b)^n$ is: $$T_{r+1} = ({\table n;r})a^{n-r}b^r$$

where $({\table n;r}) = {_n}C_r$.

The fifth row of Pascal's triangle is: 1, 5, 10, 10, 5, 1.

The binomial expansion of $(x+3/x)^5$ is therefore:

$1x^5 + 5(x^4)(3/x)^1 + 10(x^3)(3/x)^2 + 10(x^2)(3/x)^3 + 5(x^1)(3/x)^4 + 1(x^0)(3/x)^5$

$= x^5 + 15x^3 + 90x + {270}/x + {405}/{x^3} + 243/{x^5}$

The General Binomial Theorem may be used to quickly find the coefficient of a specific x term.

For example, the coefficient of $x^3$ in the expansion of $(2x+4)^6$:

$(2x+4)^6$: $a= 2x$, $b=4$, $n=6$

$T_{r+1} = ({\table n;r})a^{n-r}b^r$

$n-r=3$, so $r=n-3=6-3=3$

$T_{4} = ({\table 6;3})(2x)^{3}4^3 = {6!}/{3!3!}2^3x^34^3 = (20)⋅8⋅64x^3 = 10,240x^3$

The probability distribution of the random variable $X$ (number of successful outcomes from $n$ Bernoulli trials) is:

$$P(X=r) = (\table n;r ) p^rq^{n-r}$$$X$ follows a binomial distribution with parameters $n$ and $p$, where $X∼B(n,p)$. The third parameter is $q=1-p$.

The mode of $X$ is where the function has a maximum.

The median of $X$ = $m = {x_1 + x_2}/2$, where $x_1$ is the maximum value for which $F(x_1) ≤ 1/2$ and $x_2$ the minimum value for which $F(x_2) ≥ 1/2$.

If $X ∼ B(n, p)$ then $E(X) = μ = np$ and Var(X) = $σ^2 = npq$, where $q=1-p$.

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