Chapter 11.4, Problem 45HP

To determine

Describing how binomial distributions are connected to Pascal’s triangle.

Answer to Problem 45HP

  $\boxed{C{}_r \text{is common in both}.}$

Explanation of Solution

Given information:

Describing how binomial distributions are connected to Pascal’s triangle.

Calculation:

Consider the first $5$ rows of the Pascal’s triangle.

  $\begin{array}{ccccccccc}& & & & C{}_0& & & & \\ & & & C{}_0& & C{}_1& & & \\ & & C{}_0& & C{}_1& & C{}_2& & \\ & C{}_0& & C{}_1& & C{}_2& & C{}_3& \\ C{}_0& & C{}_1& & C{}_0& & C{}_3& & C{}_4\end{array}$

The probability of $k$ successes in $n$ trials of binomial experiment,

  $P\left(k\right)=C{}_k{\left(p\right)}^k{\left(q\right)}^{n-k}$

  $P$ represents probability of success and $q$ represents probability of failure.

  $\begin{array}{cccc}0\text{ success}& \text{1 success}& \text{2 success}& \text{n success}\\ C{}_0{\left(p\right)}^0{\left(q\right)}^{n-0}& C{}_1{\left(p\right)}^1{\left(q\right)}^{n-1}& C{}_2{\left(p\right)}^2{\left(q\right)}^{n-2}& C{}_n{\left(p\right)}^n{\left(q\right)}^{n-n}\end{array}$

The above expression is derived from the binomial expansion which is given as,

  ${\left(a+b\right)}^n=C{}_0{\left(a\right)}^0{\left(b\right)}^{n-0}+C{}_1{\left(a\right)}^1{\left(b\right)}^{n-1}+C{}_2{\left(a\right)}^2{\left(b\right)}^{n-2}+\mathrm{...}+C{}_n{\left(p\right)}^n{\left(q\right)}^{n-n}$

Well, $C{}_r$

  $r$ ranging from $0$ to $n$ is common in both Pascal’s triangle and Binomial probability distribution.

Hence $\boxed{C{}_r \text{is common in both}.}$