Question

Click it or else From police records, it has been determined that $15\%$of drivers stopped for routine license checks are not wearing seat belts. If a police officer stops $10$vehicles, how likely is it that two consecutive drivers won’t be wearing their seat belts?

(a) Describe the design of a simulation to estimate this probability. Explain clearly how you will use the partial table of random digits below to carry out your simulation.

(b) Carry out three repetitions of the simulation. Copy the random digits below onto your paper. Then mark on or directly above the table to show your results.

Short answer

Part (a) At least $2$ consecutive drivers not wearing a seat belts.

Part (b) P(at least two)=$33.33\%$

Step-by-step solution

Part (a) Step 1. Given Information 

Given that $15\%$ of drivers stopped for routine licensing checks are not wearing seat belts, it's understandable. We must determine how likely it is that two consecutive drivers will not be wearing their seat belts if a police officer stops ten automobiles.

Part (a) Step 2. Explanation

Select D as the row table. Choose the first two-digit number from table D's selected row. If the number is between 00 and 14, the person is not wearing a seat belt; if the number is greater than 14, the person is wearing a seat belt. Carry on with the next nine two-digit digits in the same way (Repetitions of the number is allowed). Calculate the amount of people who aren't wearing their seatbelts. Determine the fraction of simulations with at least two consecutive drivers without wearing a seat belt by repeating the process many times.

Part (b) Step 1. Concept

$Probability=\frac{Favorableoutcomes}{Totaloutcomes}$

Part (b) Step 2. Calculation

Determine the proportion of simulations with at least two consecutive drivers not wearing seat belts by repeating the scenario many times.

FIRST REPETITION

$$\begin{gathered} 29\Rightarrow Wearingaseatbelt \\ 07\Rightarrow Notwearingaseatbelt \\ 70\Rightarrow Wearingaseatbelt \\ 48\Rightarrow Wearingaseatbelt \\ 63\Rightarrow Wearingaseatbelt \\ 31\Rightarrow Wearingaseatbelt \\ 34\Rightarrow Wearingaseatbelt \\ 70\Rightarrow Wearingaseatbelt \\ 52\Rightarrow Wearingaseatbelt \end{gathered}$$

As a result, one out of every ten people is not wearing a seat belt for the first repetition.

SECOND REPETITION

$$\begin{gathered} 62\Rightarrow Wearingaseatbelt \\ 22\Rightarrow Wearingaseatbelt \\ 45\Rightarrow Wearingaseatbelt \\ 10\Rightarrow Wearingaseatbelt \\ 25\Rightarrow Wearingaseatbelt \\ 95\Rightarrow Wearingaseatbelt \\ 05\Rightarrow Notwearingaseatbelt \\ 29\Rightarrow Wearingaseatbelt \\ 09\Rightarrow Wearingaseatbelt \\ 08\Rightarrow Wearingaseatbelt \end{gathered}$$

As a result, in the second iteration, four out of ten people are not wearing a seat belt.

THIRD REPETITION

$$\begin{gathered} 73\Rightarrow Wearingaseatbelt \\ 59\Rightarrow Wearingaseatbelt \\ 27\Rightarrow Wearingaseatbelt \\ 51\Rightarrow Wearingaseatbelt \\ 86\Rightarrow Wearingaseatbelt \\ 87\Rightarrow Wearingaseatbelt \\ 13\Rightarrow Notwearingaseatbelt \\ 69\Rightarrow Wearingaseatbelt \\ 57\Rightarrow Wearingaseatbelt \\ 61\Rightarrow Wearingaseatbelt \end{gathered}$$

As a result, one out of every ten people is not wearing a seat belt in the third repeat. Then it's pointed out that one in every three repetitions has at least two people who aren't wearing repeats. The number of favorable outcomes divided by the total number of possible outcomes equals probability.

$P\left(atleasttwo\right)=\frac{Favourableoutcomes}{Totaloutcomes}=\frac{1}{3}=0.3333=33.33\%$