Chapter 12, Problem 12.4AYK

(a)

To determine

To find:

Theprobability that exactly $3$ blood type $O$ of the given data.

Explanation of Solution

Given:

  $\begin{array}{l}m=15\\ p=0.45\\ q=0.55\end{array}$

Concept used:

Formula

The binomial distribution

  $p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

Calculation:

Using by the binomial distribution

  $p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

  $\begin{array}{l}p\left[X=0\right]=\left(\begin{array}{l}15\\ 0\end{array}\right){\left(0.45\right)}^0{\left(0.55\right)}^{15-0}\\ p\left[X=0\right]=0.0001\\ p\left[X=1\right]=\left(\begin{array}{l}15\\ 1\end{array}\right){\left(0.45\right)}^1{\left(0.55\right)}^{14}\\ p\left[X=1\right]=0.0016\\ p\left[X=2\right]=\left(\begin{array}{l}15\\ 2\end{array}\right){\left(0.45\right)}^2{\left(0.55\right)}^{13}\\ p\left[X=2\right]=0.0090\\ p\left[X=3\right]=\left(\begin{array}{l}15\\ 3\end{array}\right){\left(0.45\right)}^3{\left(0.55\right)}^{12}\\ p\left[X=3\right]=0.0318\end{array}$

(b)

To determine

To find:

The probability that $3$ or fewer is blood type $O$ of the given data.

Explanation of Solution

Given:

  $\begin{array}{l}m=15\\ p=0.45\\ q=0.55\end{array}$

Concept used:

Formula

The binomial distribution

  $p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

Calculation:

Using by the binomial distribution

  $p\left[X=x\right]=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{q}^{n-x}$

  $\begin{array}{l}p\left[X=0\right]=\left(\begin{array}{l}15\\ 0\end{array}\right){\left(0.45\right)}^0{\left(0.55\right)}^{15-0}\\ p\left[X=0\right]=0.0001\\ p\left[X=1\right]=\left(\begin{array}{l}15\\ 1\end{array}\right){\left(0.45\right)}^1{\left(0.55\right)}^{14}\\ p\left[X=1\right]=0.0016\\ p\left[X=2\right]=\left(\begin{array}{l}15\\ 2\end{array}\right){\left(0.45\right)}^2{\left(0.55\right)}^{13}\\ p\left[X=2\right]=0.0090\\ p\left[X=3\right]=\left(\begin{array}{l}15\\ 3\end{array}\right){\left(0.45\right)}^3{\left(0.55\right)}^{12}\\ p\left[X=3\right]=0.0318\end{array}$

  $\begin{array}{l}p\left[X\le 3\right]=p\left[X=0\right]+p\left[X=1\right]+p\left[X=2\right]+p\left[X=3\right]\\ p\left[X\le 3\right]=0.0425\end{array}$