1. Describe four probability situations that involve two actions. Describe the outcomes for these situations.
Situation 1: A six-sided die with 6 equally likely sections: 1, 2, 3, 4, 5, and 6 and a fair coin with only two fair possible outcomes: heads or tails. Out of both these events both of the outcomes are equally likely of occurring. Overall, there are 12 possible outcomes, breaking this down-- there are only 6 outcomes for the die and 2 outcomes for the coin, so 6*2=12 possible outcomes. The 12 outcomes are:
1. (heads 1)
2. (heads 2)
3. (heads 3)
4. (heads 4)
5. (heads 5)
6. (heads 6)
7. (tails 1)
8. (tails 2)
9. (tails 3)
10. (tails 4)
11. (tails 5)
12. (tails 6)
Situation 2: There is a spinner with 4 equally likely sections: red, orange, yellow, and green. There is also a bag of marbles containing one blue, one purple, and one pink marble of all equally proportions. There are again 12 different possible outcomes-- breaking this down, there are 4 possible outcomes on the spinner and 3 possible outcomes in the bag, so 3*4=12 possible outcomes. The 12 outcomes are:
1. (red, blue)
2. (red, purple)
3. (red, pink)
4. (orange, blue)
5. (orange, purple)
6. (orange, pink)
7. (yellow, blue)
8. (yellow, purple)
9. (yellow, pink)
10. (green, blue)
11. (green, purple)
12. (green, pink)
Situation 3: There is a spinner with two equal parts: numbered 1 and 2. There is also a fair two sided coin that has 2 possible outcomes: heads or tails. Therefore are only 4 possible outcomes-- breaking this down, there are only two possible outcomes for the spinner and only two possible outcomes for the coin, so 2*2=4 possible outcomes. The 4 possible outcomes are:
1. (1, heads)
2. (1, tails)
3. (2, heads)
4. (2, tails)
Situation 4: There is a bag containing only 4 marbles with equally proportions: blue, yellow, green, and red. There is also a fair coin with only 2 possible outcomes: heads or tails. Therefore, there are only 8 possible outcomes-- breaking this down, there are only 4 possible outcomes in the bad filled with marbles and only two possible outcomes for the coin, therefore 4*2=8 possible outcomes. The 8 possible outcomes are:
1. (blue, heads)
2. (blue, tails)
3. (yellow, heads)
4. (yellow, tails)
5. (green, heads)
6. (green, tails)
7. (red, heads)
8. (red, tails)
2. You can use an area model or a simulation to determine the probability of a situation that involves two actions. Explain how each of these is used.
Area Model: Basically, you simply find the theoretical or experimental probability of an outcome happening. To make an area model you first start drawing a large square, preferably a 10x10 square on grid paper as it's easier to do the calculations needed later on. After that, you have to represent each outcome individually, on adjacent parts of the grid/square. You would then label these parts after the possible outcomes/sample spaces that could occur in your experiment. To represent the probabilities of your outcomes occurring this you would draw in the lines needed. In an area model, dividing a box into 4 equal sections only shows that the outcomes are equally likely, but it's not likely that your outcomes in your situation will always be equally likely so, you should take into consideration the space of the sections within your area model as it is extremely crucial to the overall process. For example, if you were flipping a coin, the area model would look like this. Shown in the picture are the different outcomes possible. The "heads+tails" written on the top represent the first flip. The "heads+tails" written on the left side represents the second flip. You can see in the area model that all the boxes are the same size, which shows that you have an equal chance of landing on any of those combinations.
Simulation: A simulation is just a way to kind of copy or simulate your actual experiment so you could get an idea of all the outcomes. You could make a simulation digitally as the experiments are conducted quicker online and within an instant your results are created accurately. There are many websites that allow you to plug in the events in a situation and the parameters to make an easy simulation. Websites such as is really helpful and is easily useable:
http://media.pearsoncmg.com/curriculum/math/cmp3/math_tools/
3. Describe how you would calculate the expected value for a probability situation.
To find the expected value for a probability situation, you first have to find the theoretical probability of each possible outcome. After that, you have to place that outcome out of a certain amount, creating faction. It's best to have your fraction out of a 100 times because then the calculations will become easier. For example, suppose we are in a one-to-one free throw situation in basketball. The question that has to be asked is, how many points will a player be expected to score out of one hundred throws? If the basketball players had a free throw accuracy of 20%, the theoretical probabilities of scoring 0,1, and 2 points would be 80%, 16%, and 4%. So, that means that out of one hundred free-throws, this player would be expected to score 24 points as 80x0=0, 16x1=16, and 4x2=8, 0+16+8=24 points. However, to find the expected value, you need to figure out the amount of points that will be earned out of one throw. So, 24/100= 0.24/1. Out of 1 throw or situation, this player would be expected to score 0.24 points.
4. Expected value is sometimes called the long-term average. Explain why this makes sense.
The term "expected value" is from the value that is expected to occur in the future, or over the long term. So, long term average is another way of saying the future, and the average is also a way of saying expected value, so they go hand in hand. Again, in many respects, long-term averages are exactly the same things as expected values as they are both terms that describe what will happen over a certain amount of times in the long-run. This is why calling expected values long-term averages makes complete sense. The expected value/long term average of an outcome will always stay the same. An average never changes unless you add another factor into the game.
Situation 1: A six-sided die with 6 equally likely sections: 1, 2, 3, 4, 5, and 6 and a fair coin with only two fair possible outcomes: heads or tails. Out of both these events both of the outcomes are equally likely of occurring. Overall, there are 12 possible outcomes, breaking this down-- there are only 6 outcomes for the die and 2 outcomes for the coin, so 6*2=12 possible outcomes. The 12 outcomes are:
1. (heads 1)
2. (heads 2)
3. (heads 3)
4. (heads 4)
5. (heads 5)
6. (heads 6)
7. (tails 1)
8. (tails 2)
9. (tails 3)
10. (tails 4)
11. (tails 5)
12. (tails 6)
Situation 2: There is a spinner with 4 equally likely sections: red, orange, yellow, and green. There is also a bag of marbles containing one blue, one purple, and one pink marble of all equally proportions. There are again 12 different possible outcomes-- breaking this down, there are 4 possible outcomes on the spinner and 3 possible outcomes in the bag, so 3*4=12 possible outcomes. The 12 outcomes are:
1. (red, blue)
2. (red, purple)
3. (red, pink)
4. (orange, blue)
5. (orange, purple)
6. (orange, pink)
7. (yellow, blue)
8. (yellow, purple)
9. (yellow, pink)
10. (green, blue)
11. (green, purple)
12. (green, pink)
Situation 3: There is a spinner with two equal parts: numbered 1 and 2. There is also a fair two sided coin that has 2 possible outcomes: heads or tails. Therefore are only 4 possible outcomes-- breaking this down, there are only two possible outcomes for the spinner and only two possible outcomes for the coin, so 2*2=4 possible outcomes. The 4 possible outcomes are:
1. (1, heads)
2. (1, tails)
3. (2, heads)
4. (2, tails)
Situation 4: There is a bag containing only 4 marbles with equally proportions: blue, yellow, green, and red. There is also a fair coin with only 2 possible outcomes: heads or tails. Therefore, there are only 8 possible outcomes-- breaking this down, there are only 4 possible outcomes in the bad filled with marbles and only two possible outcomes for the coin, therefore 4*2=8 possible outcomes. The 8 possible outcomes are:
1. (blue, heads)
2. (blue, tails)
3. (yellow, heads)
4. (yellow, tails)
5. (green, heads)
6. (green, tails)
7. (red, heads)
8. (red, tails)
2. You can use an area model or a simulation to determine the probability of a situation that involves two actions. Explain how each of these is used.
Area Model: Basically, you simply find the theoretical or experimental probability of an outcome happening. To make an area model you first start drawing a large square, preferably a 10x10 square on grid paper as it's easier to do the calculations needed later on. After that, you have to represent each outcome individually, on adjacent parts of the grid/square. You would then label these parts after the possible outcomes/sample spaces that could occur in your experiment. To represent the probabilities of your outcomes occurring this you would draw in the lines needed. In an area model, dividing a box into 4 equal sections only shows that the outcomes are equally likely, but it's not likely that your outcomes in your situation will always be equally likely so, you should take into consideration the space of the sections within your area model as it is extremely crucial to the overall process. For example, if you were flipping a coin, the area model would look like this. Shown in the picture are the different outcomes possible. The "heads+tails" written on the top represent the first flip. The "heads+tails" written on the left side represents the second flip. You can see in the area model that all the boxes are the same size, which shows that you have an equal chance of landing on any of those combinations.Simulation: A simulation is just a way to kind of copy or simulate your actual experiment so you could get an idea of all the outcomes. You could make a simulation digitally as the experiments are conducted quicker online and within an instant your results are created accurately. There are many websites that allow you to plug in the events in a situation and the parameters to make an easy simulation. Websites such as is really helpful and is easily useable:
http://media.pearsoncmg.com/curriculum/math/cmp3/math_tools/
3. Describe how you would calculate the expected value for a probability situation.
To find the expected value for a probability situation, you first have to find the theoretical probability of each possible outcome. After that, you have to place that outcome out of a certain amount, creating faction. It's best to have your fraction out of a 100 times because then the calculations will become easier. For example, suppose we are in a one-to-one free throw situation in basketball. The question that has to be asked is, how many points will a player be expected to score out of one hundred throws? If the basketball players had a free throw accuracy of 20%, the theoretical probabilities of scoring 0,1, and 2 points would be 80%, 16%, and 4%. So, that means that out of one hundred free-throws, this player would be expected to score 24 points as 80x0=0, 16x1=16, and 4x2=8, 0+16+8=24 points. However, to find the expected value, you need to figure out the amount of points that will be earned out of one throw. So, 24/100= 0.24/1. Out of 1 throw or situation, this player would be expected to score 0.24 points.
4. Expected value is sometimes called the long-term average. Explain why this makes sense.
The term "expected value" is from the value that is expected to occur in the future, or over the long term. So, long term average is another way of saying the future, and the average is also a way of saying expected value, so they go hand in hand. Again, in many respects, long-term averages are exactly the same things as expected values as they are both terms that describe what will happen over a certain amount of times in the long-run. This is why calling expected values long-term averages makes complete sense. The expected value/long term average of an outcome will always stay the same. An average never changes unless you add another factor into the game.
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