Q. 109 - Precalculus Enhanced with Graphing Utilities

Question

Between $5:00$ pm and $6:00$ pm, cars arrive at McDonald’s drive-thru at the rate of $20$ cars per hour. The following formula from probability can be used to determine the probability that x cars will arrive between $5:00$ pm and $6:00$ pm.

$P\left(x\right)=\frac{{20}^x{e}^{-20}}{x!}$

where, $x!=x\times \left(x-1\right)\times \left(x-2\right).....3\times 2\times 1$

Part (a): Determine the probability that $x = 15$ cars will arrive between $5:00$ pm and $6:00$pm.

Part (b) Determine the probability that $x = 20$ cars will arrive between $5:00$ pm and $6:00$ pm.

Step-by-Step Solution

Answer

Part (a): The probability that $x=15$ cars will arrive between $5:00pm,6:00pm$ is $0.0516$.

Part (b): The probability that $x=20$ cars will arrive between $5:00pm, 6:00pm$ is $0.0888$.

Part (a) Step 1. Determine the probability for 15 cars.

Consider the given function,

$$\begin{gathered} P\left(x\right)=\frac{{20}^x{e}^{-20}}{x!}, \\ x=15 \end{gathered}$$

Substitute the value in the given function,

$$\begin{gathered} P\left(x\right)=\frac{{20}^{15}{e}^{-20}}{15!} \\ =\frac{\left(3.2768\times {10}^{19}\right)\left(2.06115\times {10}^{-9}\right)}{15!} \\ =\frac{\left(3.2768\times {10}^{19}\right)\left(2.06115\times {10}^{-9}\right)}{1.3076\times {10}^{12}} \\ =5.1674\times {10}^{-2} \\ =0.0516 \end{gathered}$$

Part (a) Step 1. Determine the probability for 20 cars.

Consider the given function,

$$\begin{gathered} P\left(x\right)=\frac{{20}^x{e}^{-20}}{x!}, \\ x=20 \end{gathered}$$

Substitute the value in the given function,

$$\begin{gathered} P\left(20\right)=\frac{{20}^{20}{e}^{-20}}{x!}, \\ =\frac{\left(1.048576\times {10}^{26}\right)}{20!} \\ =\frac{\left(1.048576\times {10}^{26}\right)}{2.4329\times {10}^{18}} \\ =0.888\times {10}^{-1} \\ =0.0888 \end{gathered}$$