Question

Show how the derivation of the binomial probabilities $P\{X=i\}=\left(\begin{array}{c}n\\ i\end{array}\right)p^i(1-p{)}^{n-i},i=0,\dots ,n$leads to a proof of the binomial theorem $(x+y{)}^n=\sum _{i=0}^n\left(\begin{array}{l}n\\ i\end{array}\right)x^i{y}^{n-i}$when $x$and $y$are nonnegative.

Hint: Let $p=\frac{x}{x+y}$.

Short answer

Assume the hint and the point that the totality of all possible probabilities must equal one.

Step-by-step solution

Given information

Given that,

The derivation of the binomial probabilities

$P\{X=i\}=\left(\begin{array}{c}n\\ i\end{array}\right)p^i(1-p{)}^{n-i},i=0,\dots ,n$

Calculation

Define random variable $X$that is Binomial with parameters $n$and $p=\frac{x}{x+y}$. We know that the totality of all possible probabilities is equal to one, i.e.

$1=\sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right)p^k(1-p{)}^{n-k}$

Substitute $p=\frac{x}{x+y}$and $1-p=\frac{y}{x+y}$to obtain that

$1=\sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right){\left(\frac{x}{x+y}\right)}^k{\left(\frac{y}{x+y}\right)}^{n-k}$

Assume out $(x+y)$in the nominator out of the totality to obtain that

$(x+y{)}^n=\sum _{k=0}^n\left(\begin{array}{l}n\\ k\end{array}\right)x^k{y}^{n-k}$

So we have proved the binomial theorem.

Final answer

We have proved the binomial theorem.

Assume the hint and the point that the totality of all possible probabilities must equal one.