Chapter 5.1, Problem 13E

(a)

To determine

To describe how you would use slips of paper device to perform one trial of the simulation.

Explanation of Solution

In the question there is a certain archery competition and there is a $0.60$ probability of hitting the centre of the largest on each shot which corresponds with about $\text{3 out of every 5}$ shots. As,

  $\begin{array}{l}0.60\\ =\frac{60}{100}\\ =\frac{3}{5}\end{array}$

Now we will write numbers $\text{1,2,3,4 and 5}$ each on different slips of papers. And then,

  $\begin{array}{l}1,2,3=\text{Hit centre}\\ \text{4,5}=\text{Do not hit centre}\end{array}$

Then we lie the papers on a table with the numbers facing down. And then mix the papers until you no longer know which number is where on the table. And pick one of the slips of paper. If the number is $\text{1,2 or 3}$ then the shot hit the centre else the shot did not hit the centre. Now we will repeat until Quinn missed the centre twice. And also we keep track of the number of trials required until two shots did not hit the centre.

(b)

To determine

To describe how you would use random digit table device to perform one trial of the simulation.

Explanation of Solution

In the question there is a certain archery competition and there is a $0.60$ probability of hitting the centre of the largest on each shot which corresponds with about $\text{3 out of every 5}$ shots. As,

  $\begin{array}{l}0.60\\ =\frac{60}{100}\\ =\frac{3}{5}\end{array}$

Now we choose a row of the random digits table. And then select the first digits table. Then select the first digit. If the digit is between $\text{1 and 6}$ (including) then the shot will have hit the centre else the shot will not have hit the centre. So, repeat until we have obtained two shots that did to hit the centre. Thus, we keep track of the number of trials required until two shots did not hit the centre.

(c)

To determine

To describe how you would use random number generator device to perform one trial of the simulation.

Explanation of Solution

In the question there is a certain archery competition and there is a $0.60$ probability of hitting the centre of the largest on each shot which corresponds with about $\text{3 out of every 5}$ shots. As,

  $\begin{array}{l}0.60\\ =\frac{60}{100}\\ =\frac{3}{5}\end{array}$

Now we will consider numbers $\text{1,2,3,4 and 5}$ such that:

  $\begin{array}{l}1,2,3=\text{Hit centre}\\ \text{4,5}=\text{Do not hit centre}\end{array}$

Now, we enter the following command into your $\text{Ti83/Ti84}$ calculator, which will simulate $20$ shots as:

  $\text{randInt(1,5,20)}$

where randInt can be found in the MATH menu under PRB.

Thus, count the number of digits you need until you obtain two-digits that are $\text{4 or 5}$ which thus represents the number of trials needed to miss the centre twice.