Chapter 6.1, Problem 61E

Success and failure: Three components are randomly sampled, one at a time, from a large lot. As each component is selected, it is tested. If it passes the test, a success (S) occurs; if it fails the test, a failure (F) occurs. Assume that 80% of the components in the lot will succeed passing the test. Let X represent the number of successes among the three sampled components.

What are the possible values for X.

Find P(3).

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

Find P(SFS) and P(SSF).

Use the results of parts (c) and (d) to find P(2).

Find P(l).

Find P(0).

Find $\mu X.$

Find $\sigma X.$

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

Given:

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,

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

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

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.

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

Calculation:

By $FSS$ , the event of having a failure first and then the other two occurrenceare 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 $0.128$ .

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

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,

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

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

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 $0.384$ .

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

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

  $0.096$ .

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

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,

  $0.008$

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

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 $1$ .

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

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 $2.4$ .

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

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 $0.6962$