A factory owns 10 machines. On any particular day, each machine is operational with probability 0.7, independent of the status of any other machine. Let X be the random variable corresponding to the number of working machines on a particular day.

- Describe the sample space, the range of the random variable X, and calculate the PMF of X.
- Determine a numerical value for P(X >= 9).

#### Solution

A factory owns $n = 10$ machines. Each machine is operational with probability $p =0.7$, independent of the status of any other machine.

Let "O" denote that the machine is operational and "N" denote that the machine is not operational. Hence the sample space is

- The sample space is

` $$ \begin{eqnarray*} S &= & \{NNNNNNNNNN, ONNNNNNNNN, NONNNNNNNN, NNONNNNNNN,\\ & & NNNONNNNNN,NNNNONNNNN,NNNNNONNNN, NNNNNNONNN,\\ & & NNNNNNNONN, NNNNNNNNON,NNNNNNNNNO, OONNNNNNNN,\\ & & ONONNNNNNN,ONNONNNNNN,ONNNONNNNN, \cdots, OOOOOOOOOO\} \end{eqnarray*} $$ `

Let $X$ be the random variable corresponding to the number of operational machines on a particular day.

Then $X$ take the values `$X=0,1,2,\cdots,10$`

.

Here `$X\sim B(10, 0.7)$`

distribution.

The probability mass function of $X$ is

` $$ \begin{aligned} P(X=x) &= \binom{10}{x} (0.7)^x (1-0.7)^{10-x}, \\ &\quad x=0,1,\cdots, 10 \end{aligned} $$ `

- The probability that $X\geq 9$ is

` $$ \begin{eqnarray*} P(X\geq 9) & =&\sum_{x=9}^{10} P(x)\\ & =&\sum_{x=9}^{10}\binom{10}{x}(0.7)^x(1-0.7)^{10-x}\\ & =& (0.1211)+(0.0282) \\ & =& 0.1493 \end{eqnarray*} $$ `

Hence the probability that $X\geq 9$ is `$0.1493$`

.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators