Chapter 3.7, Problem 111E

a

To determine

Find the probability distributions for Y.

Answer to Problem 111E

The probability distributions for Y is,

y	0	1	2
$p\left(y\right)$	$\frac{14}{30}$	$\frac{14}{30}$	$\frac{2}{30}$

Explanation of Solution

Calculation:

A hypergeometric probability distribution of random variable Y is,

$p\left(y\right)=\frac{\left(\begin{array}{c}r\\ y\end{array}\right)\left(\begin{array}{c}N-r\\ n-y\end{array}\right)}{\left(\begin{array}{c}N\\ n\end{array}\right)};y=0,1,2,\mathrm{...},n,y\le r\text{ and }n-y\le N-r$

Let Y is defined as the number among the three that do not conform to specifications. Ten thermistors are available, and three of these are to be selected for use, and two thermistors do not conform to specifications. That is, $r=2,N=10,n=3$.

The probability distribution of Y is,

$p\left(y\right)=\frac{\left(\begin{array}{c}2\\ y\end{array}\right)\left(\begin{array}{c}10-2\\ 3-y\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}$

For $y=0$:

$\begin{array}{c}p\left(0\right)=\frac{\left(\begin{array}{c}2\\ 0\end{array}\right)\left(\begin{array}{c}10-2\\ 3-0\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}\\ =\frac{56}{120}\\ =\frac{14}{30}\end{array}$

For $y=1$:

$\begin{array}{c}p\left(1\right)=\frac{\left(\begin{array}{c}2\\ 1\end{array}\right)\left(\begin{array}{c}10-2\\ 3-1\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}\\ =\frac{2\times 28}{120}\\ =\frac{14}{30}\end{array}$

For $y=2$:

$\begin{array}{c}p\left(2\right)=\frac{\left(\begin{array}{c}2\\ 2\end{array}\right)\left(\begin{array}{c}10-2\\ 3-2\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}\\ =\frac{8}{120}\\ =\frac{2}{30}\end{array}$

Hence, the probability distributions for Y is,

y	0	1	2
$p\left(y\right)$	$\frac{14}{30}$	$\frac{14}{30}$	$\frac{2}{30}$

b

To determine

Find the probability distributions for Y.

Answer to Problem 111E

The probability distributions for Y is,

y	0	1	2	3
$p\left(y\right)$	$\frac{5}{30}$	$\frac{15}{30}$	$\frac{9}{30}$	$\frac{1}{30}$

Explanation of Solution

Calculation:

Let Y is defined as the number among the three that do not conform to specifications. Ten thermistors are available, and three of these are to be selected for use, and four thermistors do not conform to specifications. That is, $r=4,N=10,n=3$.

The probability distribution of Y is,

$p\left(y\right)=\frac{\left(\begin{array}{c}4\\ y\end{array}\right)\left(\begin{array}{c}10-4\\ 3-y\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}$

For $y=0$:

$\begin{array}{c}p\left(0\right)=\frac{\left(\begin{array}{c}4\\ 0\end{array}\right)\left(\begin{array}{c}10-4\\ 3-0\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}\\ =\frac{20}{120}\\ =\frac{5}{30}\end{array}$

For $y=1$:

$\begin{array}{c}p\left(1\right)=\frac{\left(\begin{array}{c}4\\ 1\end{array}\right)\left(\begin{array}{c}10-4\\ 3-1\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}\\ =\frac{4\times 15}{120}\\ =\frac{15}{30}\end{array}$

For $y=2$:

$\begin{array}{c}p\left(2\right)=\frac{\left(\begin{array}{c}4\\ 2\end{array}\right)\left(\begin{array}{c}10-4\\ 3-2\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}\\ =\frac{6\times 6}{120}\\ =\frac{9}{30}\end{array}$

For $y=3$:

$\begin{array}{c}p\left(3\right)=\frac{\left(\begin{array}{c}4\\ 3\end{array}\right)\left(\begin{array}{c}10-4\\ 3-3\end{array}\right)}{\left(\begin{array}{c}10\\ 3\end{array}\right)}\\ =\frac{4}{120}\\ =\frac{1}{30}\end{array}$

Hence, the probability distributions for Y is,

y	0	1	2	3
$p\left(y\right)$	$\frac{5}{30}$	$\frac{15}{30}$	$\frac{9}{30}$	$\frac{1}{30}$