Chapter 4, Problem 4.108SE

Fire Alarms A fire-detection device uses three temperature-sensitive cells acting independently of one another in such a manner that any one or more can activate the alarm. Each cell has a probability $p=.8$ of activating the alarm when the temperature reaches 135°F or higher. Let x equal the number of cells activating the alarm when the temperature reaches 135°F.

a. Find the probability distribution of x.

b. Find the probability that the alarm will function when the temperature reaches 135°F.

c. Find the expected value and the variance for the random variable x.

To determine

To find: the probability distribution for $x$ .

Answer to Problem 4.108SE

  $P\left(0\right)=0.0256;P\left(1\right)=0.1536;P\left(2\right)=0.3456;P\left(3\right)=0.3456;P\left(4\right)=0.1296$ .

Explanation of Solution

Given:

A fire detection device uses three temperature -sensitive cells.

Each cell has probability $p=0.8$ of activating the alarm reaches ${135}^{0}F$ or higher.

Calculation:

Let $A_i$ represent the house $i$ alarm is activated when temperature reaches ${135}^{0}F$ .

Therefore,

  $P\left(A_i\right)=0.8;{\forall }_i\in \left\{1,2,3\right\}$

Now, to find the probability distribution for $x$ .

The probability table is as follows:

$x$	Events associated	$p\left(x\right)$
$0$	$A_1^c\cap A_2^c\cap A_3^c$	${\left(0.2\right)}^3$
$1$	$A_1\cap A_2^c\cap A_3^c,A_1^c\cap A_2\cap A_3^c,A_1^c\cap A_2^c\cap A_3$	$3{\left(0.2\right)}^2\left(0.8\right)$
$2$	$A_1\cap A_2\cap A_3^c,A_2\cap A_2^c\cap A_3,A_1^c\cap A_2\cap A_3$	$3{\left(0.8\right)}^2\left(0.2\right)$
$3$	$A_1\cap A_2\cap A_3$	${\left(0.8\right)}^3$

Hence,the probability distribution for $x$ is $P\left(0\right)=0.0256;P\left(1\right)=0.1536;P\left(2\right)=0.3456;P\left(3\right)=0.3456;P\left(4\right)=0.1296$

To determine

To find: the probability that the alarm will function when the temperature reaches ${135}^{0}F$ .

Answer to Problem 4.108SE

  $P\left(x\ge 1\right)=0.992$ .

Explanation of Solution

Given:

A fire detection device uses three temperature -sensitive cells.

Each cell has probability $p=0.8$ of activating the alarm reaches ${135}^{0}F$ or higher.

Calculation:

Let $A_i$ represent the house $i$ alarm is activated when temperature reaches ${135}^{0}F$ .

Therefore,

  $P\left(A_i\right)=0.8;{\forall }_i\in \left\{1,2,3\right\}$

Probability that the alarm will function when temperature reaches ${135}^{0}F$ is given by $P\left(x\ge 1\right)$

  $\begin{array}{l}P\left(x\ge 1\right)=1-P\left(x\text{at 0}\right)\\ P\left(x\ge 1\right)=1-{\left(0.2\right)}^3\\ P\left(x\ge 1\right)=0.992\end{array}$

Hence, the probability that the alarm will function when the temperature reaches ${135}^{0}F$ is $P\left(x\ge 1\right)=0.992$ .

To determine

To find: the expected value and the variable for the random variable $x$ .

Answer to Problem 4.108SE

  $\mu =2.4;{\sigma }^2=0.48$ .

Explanation of Solution

Given:

A fire detection device uses three temperature -sensitive cells.

Each cell has probability $p=0.8$ of activating the alarm reaches ${135}^{0}F$ or higher.

Calculation:

Let $A_i$ represent the house $i$ alarm is activated when temperature reaches ${135}^{0}F$ .

Therefore,

  $P\left(A_i\right)=0.8;{\forall }_i\in \left\{1,2,3\right\}$

Probability that the alarm will function when temperature reaches ${135}^{0}F$ is given by $P\left(x\ge 1\right)$

  $\begin{array}{l}P\left(x\ge 1\right)=1-P\left(x\text{at 0}\right)\\ P\left(x\ge 1\right)=1-{\left(0.2\right)}^3\\ P\left(x\ge 1\right)=0.992\end{array}$

Hence, the probability distribution for $x$ is $P\left(0\right)=0.0256;P\left(1\right)=0.1536;P\left(2\right)=0.3456;P\left(3\right)=0.3456;P\left(4\right)=0.1296$

  $\begin{array}{l}\text{Mean, }\mu ={\displaystyle \sum _{x}P\left(x\right)}x\\ ={\left(0.2\right)}^3\left(0\right)+3{\left(0.2\right)}^2\left(0.8\right)\left(1\right)+3{\left(0.8\right)}^2\left(0.2\right)\left(2\right)+3{\left(0.8\right)}^3\\ =2.4\end{array}$

  $\begin{array}{l}\text{Variance, }{\sigma }^2={\displaystyle \sum _{x}P\left(x\right)}{x}^2-{\sigma }^2\\ ={\left(0.2\right)}^2{\left(0\right)}^2+3{\left(0.2\right)}^2\left(0.8\right){\left(1\right)}^2+3{\left(0.8\right)}^2\left(0.2\right){\left(2\right)}^2+{\left(0.8\right)}^2{\left(2\right)}^2\left(3\right)-{\left(2.4\right)}^2\\ =0.48\end{array}$

Hence, the expected value is $\mu =2.4$ and the variable is ${\sigma }^2=0.48$ for the random variable $x$