Question

A police officer is using a radar device to check motorists’ speeds.

Prior to beginning the speed check, the officer estimates that $40$ percent of motorists will be driving more than $5$ miles per hour over the speed limit.

Assuming that the police officer’s estimate is correct, what is the probability that among $4$ randomly selected motorists, the officer will find at least $1$ motorist driving more than $5$ miles per hour over the speed limit?

Answer 1

Step-1: Given information:

The official estimates that $40$ percent of motorists will be driving more than $5$ miles per hour over the speed limit and $4$ randomly selected motorists.
That is, $\mathrm{p}=0.40$, and $\mathrm{n}=4$.
Let $\mathrm{x}$ be the number of motorists driving more than $5$ miles per hour over the speed limit.
The random variable $\mathrm{x}$ follows binomial distribution.
$$\begin{gathered} \Rightarrow \mathrm{x}~\mathrm{Bin}(\mathrm{n},\mathrm{p}) \\ x~\mathrm{Bin}(4,0.40) \end{gathered}$$
The probability mass function of Binomial distribution is,
$\Rightarrow \mathrm{P}\left(\mathrm{x}\right)=\frac{\mathrm{n}!}{\mathrm{x}!(\mathrm{n}-\mathrm{x})!}·{\mathrm{p}}^{\mathrm{x}}·(1-\mathrm{p}{)}^{\mathrm{n}-\mathrm{x}}$

Step-2: Find the probability of at least $\mathbf{1}$ motorists driving more than $\mathbf{5}$ miles per hour over the speed limit among $\mathbf{4}$.

Apply the formula for of at least $1$ motorists $4$ randomly selected motorists:

$\Rightarrow \mathrm{P}=\mathrm{P}\left(1\right)+\mathrm{P}\left(2\right)+\mathrm{P}\left(3\right)+\mathrm{P}\left(4\right)$

Step-3: Find $\mathrm{P}\left(1\right):$

Substitute $\mathrm{p}=0.40$, $\mathrm{n}=4$ and $\mathrm{x}=1$in the formula $\mathrm{P}\left(\mathrm{x}\right)=\frac{\mathrm{n}!}{\mathrm{x}!(\mathrm{n}-\mathrm{x})!}·{\mathrm{p}}^{\mathrm{x}}·(1-\mathrm{p}{)}^{\mathrm{n}-\mathrm{x}}$:
$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(1\right)=\frac{4!}{1!(4-1)!}·{\left(0.40\right)}^1.{\left(1-0.40\right)}^{4-1}\\ & =& \frac{4\times 3\times 2\times 1!}{1\times 3\times 2\times 1!}·\left(0.40\right)·(1-0.40{)}^3\\ & =& \frac{24}{6}\times \left(0.40\right)\times {\left(0.6\right)}^3\\ & =& 4\times \left(0.40\right)\times {\left(0.6\right)}^3\\ & =& 0.3456\end{array}$

Step-4: Find $\mathrm{P}\left(2\right):$
Substitute $\mathrm{p}=0.40$, $\mathrm{n}=4$ and $\mathrm{x}=2$in the formula $\mathrm{P}\left(\mathrm{x}\right)=\frac{\mathrm{n}!}{\mathrm{x}!(\mathrm{n}-\mathrm{x})!}·{\mathrm{p}}^{\mathrm{x}}·(1-\mathrm{p}{)}^{\mathrm{n}-\mathrm{x}}$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(2\right)=\frac{4!}{2!(4-2)!}·{\left(0.40\right)}^2.{\left(1-0.40\right)}^{4-2}\\ & =& \frac{4\times 3\times 2\times 1!}{2\times 1\times 2\times 1!}·\left(0.16\right)·(0.6{)}^3\\ & =& \frac{24}{4}\times \left(0.16\right)\times {\left(0.6\right)}^3\\ & =& 6\times \left(0.16\right)\times \left(0.36\right)\\ & =& 0.3456\end{array}$

Step-5: Find $\mathrm{P}\left(3\right):$

Substitute $\mathrm{p}=0.40$, $\mathrm{n}=4$ and $\mathrm{x}=3$ in the formula $\mathrm{P}\left(\mathrm{x}\right)=\frac{\mathrm{n}!}{\mathrm{x}!(\mathrm{n}-\mathrm{x})!}·{\mathrm{p}}^{\mathrm{x}}·(1-\mathrm{p}{)}^{\mathrm{n}-\mathrm{x}}$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(3\right)=\frac{4!}{3!(4-3)!}·{\left(0.40\right)}^3.{\left(1-0.40\right)}^{4-3}\\ & =& \frac{4\times 3\times 2\times 1!}{3\times 2\times 1\times 1!}·\left(0.064\right)·(0.6{)}^1\\ & =& \frac{24}{6}\times \left(0.064\right)\times \left(0.6\right)\\ & =& 4\times \left(0.064\right)\times \left(0.6\right)\\ & =& 0.1536\end{array}$

Substitute $\mathrm{p}=0.40$, $\mathrm{n}=4$ and $\mathrm{x}=4$ in the formula $\mathrm{P}\left(\mathrm{x}\right)=\frac{\mathrm{n}!}{\mathrm{x}!(\mathrm{n}-\mathrm{x})!}·{\mathrm{p}}^{\mathrm{x}}·(1-\mathrm{p}{)}^{\mathrm{n}-\mathrm{x}}$:

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}\left(4\right)=\frac{4!}{4!(4-4)!}·{\left(0.40\right)}^4.{\left(1-0.40\right)}^{4-4}\\ & =& \frac{4\times 3\times 2\times 1!}{4\times 3\times 2\times 1!\times 0!}·\left(0.0256\right)·(0.6{)}^0\\ & =& \frac{24}{24\times 1}\times \left(0.0256\right)\times \begin{array}{rcl}& & 1\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & [\end{array}\begin{array}{rcl}& & \mathrm{Expand}\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \left(0\right)\end{array}\begin{array}{rcl}& & !\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \mathrm{to}\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & 1\end{array}\begin{array}{rcl}& & ]\end{array}\\ & =& 1\times \left(0.0256\right)\times 1\\ & =& 0.0256\end{array}$

Step-6. Substitute the values in the formula for of at least $\mathbf{1}$ motorists $\mathbf{4}$ randomly selected motorists.

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}=0.3456+0.3456+0.1536+0.0256\\ & =& 0.6912+0.1792\\ & =& 0.8704\end{array}$

Hence, the probability that among $\mathbf{4}$ randomly selected motorists, at least $\mathbf{1}$ motorists driving more than $\mathbf{5}$ miles per hour over the speed limit is $\mathbf{0}\mathbf{.}\mathbf{8704}.$