Chapter 6.1, Problem 61E

(a)

To determine

To find:

The possible values that the random variable $X$ can approach.

Answer to Problem 61E

The possible values for $X$ is found to be $0,1,2$ and $3$.

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

There are three components in a selected sample.

Out of these three, no successes can be obtained. Hence, $x=0$. Also, the number of successes can be obtained as $1,2$ or $3$.

Conclusion:

Therefore, we can identify four possible values for the random variable $X$ such that,

$\begin{array}{l}x=0\\ x=1\\ x=2\\ x=3\end{array}$

(b)

To determine

To find:

The probability to have three successes.

Answer to Problem 61E

The probability $P\left(3\right)$ is found to be,

$0.512$

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

There is $80\%$ probability to a randomly selected component to be a success one.

$P\left(S\right)=80\%=0.8$

Having three successes $\left(x=3\right)$ is denoted by $SSS$. The probability of all three selected components are successes, can be obtained by,

$\begin{array}{c}P\left(SSS\right)=P\left(S\right)\times P\left(S\right)\times P\left(S\right)\\ =0.8\times 0.8\times 0.8\\ P\left(SSS\right)=0.512\end{array}$

Conclusion:

To have three successes in the test have $0.512$ probability.

(c)

To determine

To find:

The probability that denoted by $P\left(FSS\right)$.

Answer to Problem 61E

The probability of the vent $FSS$ is calculated as,

$0.128$

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

By $FSS$, the event of having a failure first and then the other two occurrence are successes is denoted.

Probability of having a success from the lot is given as $P\left(S\right)=0.8$. Since the total probability is $1$, the probability to have a failure should be,

$\begin{array}{c}P\left(F\right)=1-P\left(S\right)\\ =1-0.8\\ P\left(F\right)=0.2\end{array}$

Therefore, the probability of $FSS$ can be obtained by the following multiplication.

$\begin{array}{c}P\left(FSS\right)=P\left(F\right)\times P\left(S\right)\times P\left(S\right)\\ =0.2\times 0.8\times 0.8\\ P\left(FSS\right)=0.128\end{array}$

Conclusion:

The probability of the event $FSS$ is found to be $0.128$.

(d)

To determine

To find:

The probabilities of the events $SFS$ and $SSF$.

Answer to Problem 61E

The probabilities of the events $SFS$ and $SSF$ are found to be $0.128$ for both.

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

As calculated in the part (c),

$P\left(S\right)=0.8$ and $P\left(F\right)=0.2$

Hence, the probability of the event $SFS$ should be,

$\begin{array}{c}P\left(SFS\right)=P\left(S\right)\times P\left(F\right)\times P\left(S\right)\\ =0.8\times 0.2\times 0.8\\ P\left(SFS\right)=0.128\end{array}$

Also, the probability of the event $SSF$ should be,

$\begin{array}{c}P\left(SSF\right)=P\left(S\right)\times P\left(S\right)\times P\left(F\right)\\ =0.8\times 0.8\times 0.2\\ P\left(SSF\right)=0.128\end{array}$

Conclusion:

The probabilities are found to be,

$\begin{array}{c}P\left(SFS\right)=0.128\\ P\left(SSF\right)=0.128\end{array}$

(e)

To determine

To find:

The value of $P\left(2\right)$ by $P\left(FSS\right),P\left(SFS\right)$ and $P\left(SSF\right)$.

Answer to Problem 61E

The probability is calculated as,

$P\left(2\right)=0.384$

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

Occurring two successes is represented by $x=2$ where $X$ is the random variable of number of successes.

There are three different ways that two successes such that $SSF,SFS$ and $FSS$. The corresponding probabilities are found to be as follows.

$\begin{array}{c}P\left(FSS\right)=0.128\\ P\left(SFS\right)=0.128\\ P\left(SSF\right)=0.128\end{array}$

Therefore, the probability $P\left(2\right)$ can be obtained by the addition of the above three values.

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

Conclusion:

The probability to have two successes is $0.384$.

(f)

To determine

To calculate:

The probability to obtain one success.

Answer to Problem 61E

The probability is found to be,

$P\left(1\right)=0.096$

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

Within all the possible results of the test, there are only three combinations that have only one success.

$SFF,FSF$ and $FFS$

Each combination has two failures and one success. Hence, the probability have anyone of these three is equal and it can be calculated as,

$\begin{array}{c}P=0.2\times 0.2\times 0.8\\ =0.032\end{array}$

The multiplication of these three probabilities and three returns the probability to have one success within the test.

$\begin{array}{c}P\left(1\right)=3\times 0.032\\ =0.096\end{array}$

Conclusion:

The probability for obtaining one success is found to be $0.096$.

(g)

To determine

To find:

The probability to not have any successes.

Answer to Problem 61E

The probability is found to be,

$P\left(0\right)=0.008$

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

To obtain no success, the selected three components should be failures. Hence, there is only one combination for that event.

Selecting three failures from the lot is denoted by $P\left(FFF\right)$ and the probability should be,

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

Conclusion:

The probability to have no successes is found to be,

$0.008$

(h)

To determine

To find:

The mean of the number of successes.

Answer to Problem 61E

The mean is found to be,

${\mu }_X=2.4$

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

The probabilities of $X=\left\{0,1,2,3\right\}$ is said to be $0.008,0.096,0.384$ and $0.512$ respectively. These are the all possibilities of the random variable $X$ and it can be verified by the total probability. The sum of probabilities of all possible outcomes should be $1$.

$\begin{array}{c}P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)=0.008+0.096+0.384+0.512\\ =1\end{array}$

The mean of the random variable is given by the sum of the values of random variables and the corresponding probabilities.

$\begin{array}{c}{\mu }_X=\left(0\times 0.008\right)+\left(1\times 0.096\right)+\left(2\times 0.384\right)+\left(3\times 0.512\right)\\ =0+0.096+0.768+1.536\\ {\mu }_X=2.4\end{array}$

Conclusion:

The mean of $X$ is calculated as $2.4$.

(i)

To determine

To find:

The standard deviation of the number of successes.

Answer to Problem 61E

The standard deviation is calculated as,

${\sigma }_X=0.6962$

Explanation of Solution

In the test three components from a lot which has $80\%$ success rate, are randomly selected and tested. An occurred success is denoted by $S$ and a failure is denoted by $F$. The random variable $X$ represents the number of successes out of three components.

Calculation:

The formula to calculate the variance of a random variable $\left({\sigma }_X{}^2\right)$ is,

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

In the part (h), the mean is calculated as $2.4$. Assigning data into the formula is easy with a table.

$\begin{array}{ccccc}x& x-{\mu }_X& {\left(x-{\mu }_X\right)}^2& P\left(x\right)& {\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)\\ 0& -2.4& 5.76& 0.008& 0.04608\\ 1& -1.4& 1.96& 0.096& 0.18816\\ 2& -0.4& 0.16& 0.384& 0.06144\\ 3& 0.6& 0.36& 0.512& 0.18432\end{array}$

The sum of right-most column gives the variation of $X$.

${\sigma }_X{}^2=\mathrm{0.0.484704}$

The standard deviation $\left({\sigma }_X\right)$ is the square root of variance. Hence,

$\begin{array}{c}{\sigma }_X=\sqrt{{\sigma }_X{}^2}\\ =\sqrt{0.484704}\\ {\sigma }_X=0.6962\end{array}$

Conclusion:

The standard deviation of $X$ is said to be $0.6962$. $X$