Chapter 5.1, Problem 61E

a.

To determine

Find the possible values of X.

Answer to Problem 61E

The possible values of X are 0, 1, 2 and 3.

Explanation of Solution

From a large lot, three components are sampled, one at a time. The component is tested at that time it self and if the component passes the test it is denoted as success (S) and if the component fails the test it is considered as a failure (F). The random variable X denotes the number of success among the three components. It was found that 80% of the components succeed in passing the test.

Here, the random variable X represents the number of success among the three components. That is, the number of component that passes the test. Thus, the random variable X can take values 0, 1, 2 and 3.

b.

To determine

Find the probability, $P\left(3\right)$.

Answer to Problem 61E

The probability $P\left(3\right)$ is 0.512.

Explanation of Solution

Calculation:

Multiplication rule:

If the events A and B are independent, then $P\left(A\text{and }B\right)=P(A)\times P\left(B\right)$.

Here, a random sample of 3 components is taken. Each component is independent of the other. It was found that 80% of the components in the lot will pass the test. Thus, the probability of success (p) is 0.80.

The probability that all 3 components pass the test, $P\left(3\right)$ is the probability that the first component, second component and the third component will pass the test. That is,

$\begin{array}{c}P\left(\begin{array}{l}\text{All the three components}\\ \text{pass the test}\end{array}\right)=P\left(SSS\right)\\ =P\left(S\right)\times P\left(S\right)\times P\left(S\right)\\ =\left(0.80\right)\times \left(0.80\right)\times \left(0.80\right)\\ =0.512\end{array}$

Thus, the probability $P\left(3\right)$ is 0.512.

c.

To determine

Find the probability, $P\left(FSS\right)$.

Answer to Problem 61E

The probability $P\left(FSS\right)$ is 0.128.

Explanation of Solution

Calculation:

The event that the first component fails and the next two succeed is denoted by FSS.

The probability $P\left(FSS\right)$ is the probability that the first component will fail the test and the second component and the third component will pass the test. The probability that a component will pass the test is 0.80, the probability that a component will fail the test is, $0.20\left(=1-0.80\right)$. That is,

$\begin{array}{c}P\left(FSS\right)=P\left(F\right)\times P\left(S\right)\times P\left(S\right)\\ =\left(1-0.80\right)\times \left(0.80\right)\times \left(0.80\right)\\ =\left(0.20\right)\times \left(0.80\right)\times \left(0.80\right)\\ =0.128\end{array}$

Thus, the probability $P\left(FSS\right)$ is 0.128.

d.

To determine

Find the probability, $P\left(SFS\right)$ and $P\left(SSF\right)$.

Answer to Problem 61E

The probability $P\left(SFS\right)$ is 0.128.

The probability $P\left(SSF\right)$ is 0.128.

Explanation of Solution

Calculation:

The event that the second component fails and the first and third succeed is denoted by SFS.

That is, the probability $P\left(SFS\right)$ is,

$\begin{array}{c}P\left(SFS\right)=P\left(S\right)\times P\left(F\right)\times P\left(S\right)\\ =\left(0.80\right)\times \left(1-0.80\right)\times \left(0.80\right)\\ =\left(0.80\right)\times \left(0.20\right)\times \left(0.80\right)\\ =0.128\end{array}$

Thus, the probability $P\left(FSS\right)$ is 0.128.

The event that the first and second succeed and the third component fails is denoted by SSF.

That is, the probability $P\left(SSF\right)$ is,

$\begin{array}{c}P\left(SSF\right)=P\left(S\right)\times P\left(S\right)\times P\left(F\right)\\ =\left(0.80\right)\times \left(0.80\right)\times \left(1-0.80\right)\\ =\left(0.80\right)\times \left(0.80\right)\times \left(0.20\right)\\ =0.128\end{array}$

Thus, the probability $P\left(SSF\right)$ is 0.128.

e.

To determine

Find the probability, $P\left(2\right)$.

Answer to Problem 61E

The probability $P\left(2\right)$ is 0.384.

Explanation of Solution

Calculation:

The probability $P\left(2\right)$ is the probability that two components passed the test and one failed. That is,

$\begin{array}{c}P\left(2\right)=P\left(FSS\right)\text{ or }P\left(SFS\right)\text{ or }P\left(SSF\right)\\ =\left[P\left(FSS\right)\text{+}P\left(SFS\right)\text{+}P\left(SSF\right)\right]\end{array}$

Substitute the values from part (c) and (d).

Then the required probability is,

$\begin{array}{c}P\left(2\right)=\left[P\left(FSS\right)\text{+}P\left(SFS\right)\text{+}P\left(SSF\right)\right]\\ =0.128+0.128+0.128\\ =0.384\end{array}$

Thus, the probability $P\left(2\right)$ is 0.384.

f.

To determine

Find the probability, $P\left(1\right)$.

Answer to Problem 61E

The probability $P\left(1\right)$ is 0.096.

Explanation of Solution

Calculation:

The probability $P\left(1\right)$ is the probability that one component passed the test and two failed. That is,

$\begin{array}{c}P\left(1\right)=P\left(FFS\right)\text{ or }P\left(FSF\right)\text{ or }P\left(SFF\right)\\ =\left[P\left(FFS\right)\text{+}P\left(FSF\right)\text{+}P\left(SFF\right)\right]\end{array}$

The probability that a component will pass the test is 0.80, the probability that a component will fail the test is $0.20\left(=1-0.80\right)$. That is,

$\begin{array}{c}P\left(1\right)=\left[P\left(FFS\right)\text{+}P\left(FSF\right)\text{+}P\left(SFF\right)\right]\\ =\left[\left(0.2\times 0.2\times 0.8\right)+\left(0.2\times 0.8\times 0.2\right)+\left(0.8\times 0.2\times 0.2\right)\right]\\ =\left[0.032+0.032+0.032\right]\\ =0.096\end{array}$

Thus, the probability $P\left(1\right)$ is 0.096.

g.

To determine

Find the probability, $P\left(0\right)$.

Answer to Problem 61E

The probability $P\left(0\right)$ is 0.008.

Explanation of Solution

Calculation:

The probability $P\left(0\right)$ is the probability that all the components failed the test. That is,

$P\left(0\right)=P\left(FFF\right)$

The probability that a component will fail the test is $0.20\left(=1-0.80\right)$. That is,

$\begin{array}{c}P\left(0\right)=P\left(FFF\right)\\ =\left(0.2\times 0.2\times 0.2\right)\\ =0.008\end{array}$

Thus, the probability $P\left(0\right)$ is 0.008.

h.

To determine

Find the mean ${\mu }_X$.

Answer to Problem 61E

The mean value is 2.4.

Explanation of Solution

Calculation:

The formula for the mean of a discrete random variable is,

$\begin{array}{c}E\left(X\right)={\mu }_X\\ ={\displaystyle \sum x\cdot P\left(x\right)}\end{array}$

The values of the random variable X and the corresponding probabilities obtained from parts (b), (e), (f) and (g) are given in the following table:

x	P(x)
0	0.008
1	0.096
2	0.384
3	0.512

The mean of the random variable is obtained as given below:

x	P(x)	$x\cdot P\left(x\right)$
0	0.008	0.000
1	0.096	0.096
2	0.384	0.768
3	0.512	1.536
Total	1.000	2.400

Thus, the mean value ${\mu }_X$ is 2.4.

i.

To determine

Find the standard deviation ${\sigma }_X$.

Answer to Problem 61E

The standard deviation ${\sigma }_X$ is 0.6928.

Explanation of Solution

Calculation:

The standard deviation of the random variable X is obtained by taking the square root of variance.

The formula for the variance of the discrete random variable X is,

${\sigma }_X^2={\displaystyle \sum \left[{\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)\right]}$

Where ${\mu }_X={\displaystyle \sum \left[x\cdot P\left(x\right)\right]}$ represents the mean of the random variable X.

The variance of the random variable X is obtained using the following table:

x	P(x)	$\left(x-{\mu }_X\right)$	${\left(x-{\mu }_X\right)}^2$	${\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)$
0	0.008	–2.4	5.76	0.04608
1	0.096	–1.4	1.96	0.18816
2	0.384	–0.4	0.16	0.06144
3	0.512	0.6	0.36	0.18432
Total	1.000	–3.6	8.24	0.48

Therefore,

${\sigma }_X^2=0.48$

Thus, the variance is 0.48.

The standard deviation is,

$\begin{array}{c}{\sigma }_X=\sqrt{0.48}\\ =0.6928\end{array}$

That is, the standard deviation is 0.6928.