# Solved (Free): A factory owns 10 machines. On any particular day, each machine is operational with probability 0.7

#### ByDr. Raju Chaudhari

Mar 8, 2021

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.

1. Describe the sample space, the range of the random variable X, and calculate the PMF of X.
2. 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

1. 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}

1. 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 $0.1493$.