My question:
A spinner looks like this: A disc divided into sections that might or might not be the same size, with a pointer that when spun will land on some part of the disc at random. The sections might be colours, or, as in these problems, have numbers in them.
There are two problems, not associated with each other. The first is to help you get the idea. The second is more challenging. You will be given a series of six clues for each problem that should enable you to draw the spinners.
Problem 3a:
•
The four numbers on the spinner are equally likely. They are also all different.
•
The smallest sum you can get in two spins of this spinner is 2; the largest is 14.
•
It is impossible to get an odd number if you spin this spinner twice and add the results.
•
If you spin this spinner twice, and add the two numbers, you are just as likely to get 10 as 6.
•
The most likely sum of two spins of this spinner is eight.
•
It is impossible to get an even number on this spinner if you spin three times and add.
•
DRAW THE SPINNER.
Problem 3b:
•
The spinner has three sections. While no two are the same size, one of the sections is half the size of another.
•
You are more likely to get 2/3 than one if you sum two spins, but 5/6 is the most likely sum of all.
•
The sum of the three numbers on the spinner is one.
•
The largest number you can get in two spins of this spinner is one.
•
If you spin the spinner twice and add, you get a sum of one about a quarter of the time.
•
If you spun the spinner a hundred times and added up all the numbers, you'd probably get somewhere near 40.
•
DRAW THE SPINNER.
Although I cannot completely solve the problem (3a), I can give you some help.
If the highest number for two spins is 14 and the lowest is 2, one number must be 1 and the other 7.
If it is impossible to get an odd number from adding 2 spins, all the numbers have to be odd. That goes along with the final clue too.
With the other clues, it seems like the numbers are 1, 3, 5 and 7. Check to see if it fits.
I hope this helps you get the idea for solving the second problem, where the numbers are fractions. Thanks for asking.
Sorry for replying so late after you have done this but I only just came across the page.
for the first one the answer is 1,7,3,5. As the lowest number must be a 1 and the highest a 7, all numbers must be odd if they come up with an even number after two spins and odd after 3. the only two odds in between 1 and 7 are 3 & 5.
For the second one I have found an answer that explains all but the second part of 2 and number 5. I get 1/3; 1/6 & 1/2. the highest number must be 1/2 to equal to 1. As one number is half of another but not the same as any other we try to use half of 1/2 which is 1/4 but this will mean we have 2 x 1/4 and 1/2 which we can not do. 1/3 is 2x1/6 and comes up with 2/3 when adding 2 spins together on 50% of the spins.The equation we have to solve is 2x + x + y = 1 so x=1/6 and y = ½.
I hope this helps as I have never done these before.
For Problem 3a, based on the given clues, we can conclude the following:
- The spinner has four numbers that are equally likely and all different.
- The smallest sum you can get from two spins is 2, and the largest sum is 14.
- It is impossible to get an odd number by spinning the spinner twice and adding the results.
- The probability of getting a sum of 10 is the same as getting a sum of 6 when spinning the spinner twice and adding the numbers.
- The most likely sum from two spins is 8.
- It is impossible to get an even number by spinning the spinner three times and adding.
Based on these clues, we can draw the spinner as follows:
```
Spinner for Problem 3a:
____
/ \
| 2 |
\____/
/ \
| 8 |
\____/
/ \
| 4 |
\____/
/ \
| 6 |
\____/
```
For Problem 3b, based on the given clues, we can conclude the following:
- The spinner has three sections, and no two sections are the same size.
- One section is half the size of another section.
- The probability of getting 2/3 is higher than getting 1 in the sum of two spins.
- The most likely sum of all three numbers is 5/6.
- The sum of the three numbers on the spinner is 1.
- The largest number you can get from two spins is 1.
- If you spun the spinner a hundred times and added up all the numbers, you would probably get somewhere near 40.
Based on these clues, we can draw the spinner as follows:
```
Spinner for Problem 3b:
______
/ \
| 1 |
\______/
/ \
| 1/2 |
\______/
/ \
| 1/3 |
\______/
```
Please note that the sizes of the sections may not be accurately proportioned in the text representation above.
To draw the spinners for problems 3a and 3b, we need to analyze the given clues and use deductive reasoning to determine the possible configurations. Let's break down each problem and work through the clues step by step.
Problem 3a:
1. The four numbers on the spinner are equally likely and all different: This means that each number appears once, and no two numbers are the same. So, we can represent the spinner with the numbers 1, 2, 3, and 4.
2. The smallest sum you can get in two spins is 2, and the largest is 14: To achieve a sum of 2, we need the two smallest numbers on the spinner, which are 1 and 1. And to obtain a sum of 14, we need the two largest numbers, which are 4 and 4.
3. It is impossible to get an odd number if you spin the spinner twice and add the results: This means that we cannot have any odd sums. Since the smallest sum is 2 and we can't get odd sums, the only possible combinations are (1,1), (2,2), and (4,4).
4. The probability of getting a sum of 10 is the same as getting a sum of 6: This implies that there are multiple ways to obtain a sum of 10 and a sum of 6. The only possible combinations are (4,6) and (5,5).
5. The most likely sum of two spins is 8: Based on the previous clues, the possible combinations that can give us a sum of 8 are (3,5) and (4,4).
6. It is impossible to get an even number on this spinner if you spin three times and add: If we need to get an odd sum by adding three numbers, we can only use one even number. Thus, the spinner cannot have any number more than once.
Based on these clues, we can draw the spinner for problem 3a as follows:
1 2 3 4
-----------------
| |
| 4 5 6 |
| |
|-------|-------|
| 3 | 1 |
|-------|-------|
| |
| 2 8 7 |
| |
-----------------
Problem 3b:
1. The spinner has three sections, and no two are the same size: This means that the sections must have distinct sizes.
2. One section is half the size of another: Let's assume that the larger section is represented by x, then the smaller section will be x/2.
3. You are more likely to get 2/3 than one if you sum two spins: This implies that the section representing 2/3 should be larger than the one representing 1. So, let's assign the larger section as 2/3 and the smaller section as 1/3.
4. The sum of the three numbers on the spinner is one: Since the sum has to be one, we can assume that the remaining section represents -2/3 to balance out the positive values. This means that we have -2/3, 1/3, and 2/3.
5. The largest number you can get in two spins is one: By adding the largest and smallest sections, we should get 1. Using this clue, we can deduce that the largest section should be 2/3.
6. If you spin the spinner twice and add, you get a sum of one about a quarter of the time: This gives us an indication that the 2/3 section should appear more frequently. So, the remaining two sections are 1/3 and -2/3.
Based on these clues, we can draw the spinner for problem 3b as follows:
| /\
| / \
| / \
-2/3 2/3 1/3
--------------------------------
| \ / |
| 1 |
| |
--------------------------------
Note: The spinner is not perfectly symmetrical, and the sections may not be equal in size. The shapes above are just rough representations.