# I need help to work out all questions on the assignment page.

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Probability is about the chances of something happening. In the problems that are featured on the assignment page, you are looking at different situations in which the focus lies in the chances of an outcome being determined. For the tasks in the first question, we are looking at a situation in which we start off with seven red counters in a bag. From that point, the questions move into asking clarification based on terms. When looking at probability, an event is seen as "certain" when there is a 100% chance of it occurring. An event can be seen as a 50/50 when there is an even chance of the event happening. In probability terminology, "even chance" translates to a 50% chance of something happening. There is a 50% chance it can happen and a 50% chance that it does not happen. When we are looking for a probability of exactly 1 out of 3, then we are seeing to find an event that has a 33% chance of happening. When an event is impossible, that means that there is a probability that it will not happen, or we are sure it will not take place. From this, we progress into our seven red counters. There is a certain chance that a red counter will be pulled if no blue is added. This means that if no blue counters are added, then we are certain that we will reach in and pull a red counter. When we want to move to 50/50, then we look at the number of red counters we have, which is 7. To get our probability to 50%, we will end up adding 7 blue counters. This will give our total number of counters to 14 and 7 of them will be red. Expressing this as a fraction, we have the number of desired outcomes as our numerator and the number of total outcomes as our denominator. 7/14 (the number of red counters/ the total number of counters) comes out to 50%. Thus, we need to add 7 blue counters. This same philosophy is seen when we want to work a 1 out of 3 chance. If we add 14 blue counters, our total number of counters in the bag goes to 21, and with our 7 blue counters, we have a 7/21 chance of pulling a red counter, which reduces to 1/3, or 1 out of 3 chance. Thus, to get a 1 out of three probability, add 14 blue counters. Finally,from a probability standpoint, it is impossible to make the outcome of pulling a red to statistically go away. Since there is at least one red counter in the bag, it is possible to pull a red. This outcome is always going to be there. In this light, it is impossible to find an impossibility for a red counter. There will be red counters in the bag, and thus there is always a chance that red counters can be pulled.

When looking at our next problem on the page, we progress into a more complex arena of probability. However, the concepts will remain the same, even though the operations become more intricate. Recall that our formula for determining probability is setting up a fraction that shows the number of likely or successful outcomes over the total number of outcomes. In this, we can express the probability of a single event taking place. When looking at the spinner and the tallies of outcomes being 1-4, we have to look at a couple of things. The first is that we have to tally the total number of frequencies. We do this by adding everything up. When we do this, we recognize that we have a total of 60 frequencies that have been recorded in the data table on the page. When we look at the individual tallies out of the total of 60, we can start to express the probability of events happening (For purposes of showing our work, we will express the probability (P) of a particular event (number) happening:

60 total outcomes

P (1)= 15/60- 25%- There were 15 total tallies of "1" and when we put it over the total, 15/60 can be reduced to 1/4 or 25%

P (2)= 6/ 60- 10%

P (3)= 18/60- 30%

P (4)= 21/60- 35%

This means the following:

P(2)= Least likely- The "2" is the outcome that features the least amount of tallies.

P (1)= Second Least Likely

P (3)= Second Most Likely

P (4)- Most likely- The "4" is the outcome that features the most amount of tallies.

The most important element here is to add up the total number of tallies or frequencies. From being able to generate the total at 60, we can progress from there in seeing how many frequencies each option renders. In setting up a ratio of the number of desirable outcomes as the numerator and the number of total outcomes as the denominator, we can express the probability of each event happening based on the data we have collected. From this, we see that of the 60 total outcomes the Probability of drawing a 1 with the spinner, or P (1), equals 15/60. We know that there are sixty total outcomes because we add up all four sets of frequencies and get 60. From this, 15 out of the 60 were 1. P(2)= 6/60, while P(3)= 18/60 and P(4)= 21/60.

In estimating the probability, we see that the probabilities give a fairly clear percentage of breakdown. P(1) is 1 out of 4,or a 25%, while P(2) is 1 out of 10 or 10%. P(3) comes out to three out of 10, which is a 30%, while P(4) is between 3 and 4 out of 10, which can be seen as between 30 and 40% when we estimate it.

**Sources:**

3A) In order to make sure that the counter chosen is always red, the only counters that can be inside the container have to be red. Thus, the only way to be **certain** that the counter chosen is red is to not add any blue counters.

3B) In order to make it so that the chance of choosing a red counter is **50/50**, there must be an equal number of blue counters as red counters. In other words, 7/x = 1/1. In this case, you must add 7 blue counters.

3C) In order to make the chances equal **one-in-three**, the total number of counters in the bag must be three times that of the blue counters. In other words, 7/x = 1/3, in which x is the total number of counters. Solving for x gives 21. Then, we must subtract 7 from 21 since there are already 7 red counters and the problem wants to know how many blue counters to ADD. That gives us 14 blue tokens to be added.

3D) Since there are already 7 red counters in the bag, there is no way that it would be **impossible** to choose a red counter except to remove all the red counters.

4A)

I) The order is as follows: 2, 1, 3, 4.

II) In a total of 60 tries, the spinner landed on "2" only 6 times, making it the least likely. In the same way, we can arrange the frequencies in increasing order to see which number is most likely. Thus, 6, 15, 18, and 21 correspond to 2, 1, 3, and 4, respectively.

4B) This requires calculation of the total number of trials done. Since 6+15+18+21 = 60, that will be determined to be the total number of trials. Now we can simply put the frequency of each number over the total of 60.

Probability = frequency of a number/total tries

I) Probability = 15/60 = 1/4 or 25%

II) Probability = 6/60 = 1/10 or 10%

III) Probability = 18/60 = 3/10 or 30%

IV) Probability = 21/60 = 7/20 or 35%

**3a.** If you wanted it to be absolutely certain that you would get a red counter, you would put **0** blue counters in the bag. Even one blue counter would make a slight chance for you to not get red.

**3b.** If you wanted a 50/50 chance, 50% that you'd get red and 50% that you'd get blue, you would want an equal amount of red and blue counters. Since you have 7 red, you would need to add **7** blue to your bag.

**3c.** If you wanted a 1/3 chance to get a red counter, you would need to be able to split your total counters into three equal parts. The easiest way to get that is, since you already have 7 red counters, to multiply 7 by 3. You need a total of 21 counters, meaning you will need to add **14** blue counters. That will make 1/3 of the counters be red, and you'll have a 1/3 chance of getting a red counter.

**3d.** There isn't a way to make it *impossible* to draw red. If there are 7 red counters in the bag, no matter how many blue you add, there will always be a slight chance of drawing red.

**4a(i).** Your order from least likely to most likely is **2, 1, 3, 4**.

**4a(ii).** To do this, you would use your frequencies from the table. You spun a 2 six times, which was the least, and you spun a 4 twenty-one times which was the most.

**4b(i).** According to the table, you spun a total of 60 times. Since 15 of those spins landed on the number 1, your probability is 15/60 or **1/4** or 25%.

**4b(ii).** You spun the number 2 six times out of 60, which makes you probability 6/60 or **1/10** or 10%.

**4b(iii).** You spun the number 3 eighteen times, which makes your probability 18/60 or **3/10** or 30%.

**4b(iiii).** You spun the number 4 twenty-one times, which makes your probability 21/60 or **7/20** or between 30-40%.

3a. 0

3b. 7 blue for 7 red

3c. 14 blue to 7 red for 1/3 chance

3d. Impossible for impossible as there is at least 1 red in the bag.

1. 1/10

2. 1/4

3. 3/10

4. 7/20

Problem 3:

a) if you want to be certain your pulling red, than don't add any blue

b) a 50/50 chance means that they are equal, or that half the bag is red and the other blue, so add 7 blue

c) multiply 1/3 by 7/7 to get 7/21. 21 is the total in the bag so subtract the number of red (7) and you know that you need to add 14 blue

d) it is impossible to make it impossible to get red, unless you remove red entirely

Problem 4:

a) 2, 1, 3, 4

b) The numbers with the least tally's happened the least, and the ones with most happened more often making them more likely statistically.

c) a\Add the number of tally's to get 60. Then put each individual number's tally's and put over 60 (and simplify) to get the probability ratio

#1: 6/60 = 1/10

#2: 15/60 = 1/4

#3: 18/60 = 3/10

#4: 21/60 = 7/20