Chapter 5: Q. 13 (page 310)

Bull’s-eye! In a certain archery competition, each player continues to shoot until he or she misses the center of the target twice. Quinn is one of the archers in this competition. Based on past experience, she has a $0.60$ probability of hitting the center of the target on each shot. We want to design a simulation to estimate the probability that Quinn stays in the competition for at least $10$ shots. Describe how you would use each of the following chance devices to perform one trial of the simulation.
a. Slips of paper
b. Random digits table
c. Random number generator

Short Answer

Expert verified

a. The papers are then placed on a table, numbers facing down. Then jumble the sheets so you can't tell which number is on which table. And choose one of the paper slips. If the number is $1, 2,$ or $3$ , the shot hit the centre; otherwise, the shot did not.

b. We'll now pick a row from the random digits database. After that, choose the first digits table. After that, choose the first digit. The shot will have hit the centre if the digit is between $1$ and $6$ (inclusive); otherwise, the shot will not have hit the centre.

c. we enter the following command into your $T i 83 / T i 84$ calculator, which will simulate $20$ shots as:

$r a n d I n t (1, 5, 20)$

where $r a n d \ln t$ can be found in the MATH menu under PRB.

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

Step by step solution

Part (a) Step 1 : Given Information

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

Part (a) Step 2 : Simplification

There is an archery competition in the question, and each shot has a

0.60

chance of reaching the centre of the largest, which amounts to around

3

out of every

5

shots.

As,

$0.60 = \frac{60}{100} = \frac{3}{5}$

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

$1, 2, 3$ =Hit centre $4, 5$ =Do not hit centre

The papers are then placed on a table, numbers facing down. Then jumble the sheets so you can't tell which number is on which table. And choose one of the paper slips. If the number is

1, 2,

3

, the shot hit the centre; otherwise, the shot did not. We'll keep going until Quinn misses the centre twice. We also keep track of how many trials are required until two shots do not hit the centre.

Part (b) Step 1 : Given Information

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

Part (b) Step 2 : Simplification

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 $3$ out of every $5$ shots. As,

$0.60 = \frac{60}{100} = \frac{3}{5}$

We'll now pick a row from the random digits database. After that, choose the first digits table. After that, choose the first digit. The shot will have hit the centre if the digit is between

1

and

6

(inclusive); otherwise, the shot will not have hit the centre. So, keep repeating until we have two bullets that hit the centre. As a result, we keep track of the number of tries required until no more than two rounds miss the target.

Part (c) Step 1 : Given Information

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

Part (c) Step 2 : Simplification

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 $3$ out of every $5$ shots. As,

$0.60 = \frac{60}{100} = \frac{3}{5}$

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

$1, 2, 3$ =Hit centre $4, 5$ =Do not hit centre

Now, we enter the following command into your $T i 83 / T i 84$ calculator, which will simulate $20$ shots as:

$r a n d I n t (1, 5, 20)$

where $r a n d I n t$ can be found in the MATH menu under PRB. Thus, count the number of digits you need until you obtain two-digits that are $4$ or $5$ which thus represents the number of trials needed to miss the centre twice.