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We simulated 184 runs to see the outcome of how many times you would win in 15 bottles. The probability was 1 in 5 which means we should see 3 in 15 being the most common, we don't see because of the sampling variability instead we see that 2 in 15 is actually the most common.

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Out of a set of random numbers (1-15), the numbers 7, 12 and 13 represented a winning prize of chocolate milk. By chance acting alone, it is said that only receiving two bottles of chocolate milk is 'unfair" In experimental terms, picking only two of the winning numbers, contrary to the advertised theoretical probability of 0.3 is apparently unfair. Because we can't say for sure (say why this is?) whether they have a point or not we underwent a simulation of 184 sets of 15 using computer software online. This simulation enabled us to investigate their hypothesis further.

Out of a context of winning chocolate milk bottles, there was a set of 15 numbers where 7, 12 and 13

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in our simulation, we did 184 runs to see the frequency of our selected prize winning numbers (7,12,13). We did the runs in groups of 15, seeing how many times the lucky numbers occured in that group of 15. The theo

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We simulated this 184 to see the outcome of bowman many times you would win in 15 bottles. The probability we were told was it would be 1 in 5 which means we should see 3 in 15 being the most common. However we don't see because of the sampling variability. We therefore see that 2 in 15 is actually the most common.  By chance acting alone you could possibly get 0,1,2 instead of the 3 in 15 winners.

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In our simulation, we had a sample of 15 numbers ranging from 1-15 and assigned three winning numbers (7,12,13). We then simulated the tests 184 times, and recorded how many times the winning numbers were selected in each sample of 15. The true probability of the simulation was 1 in 5, so we would expect to see 3 successes in each test. The graph, however, shows the most common number of successes was 2 which does not follow the trend of the true probability. In order for the simulation to follow the same trend as the true probability (1 in 5) we would need to increase our sample size.

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We used a simulation to find groups of 15 random numbers, ranging from 1-15. From each group, we recorded the number of 'successes' we got. (i.e 7, 12 or 13) We completed a total of 184 simulations. Even though our most common frequency of scuccesses per group was slightly lower than the theoretical, due to chance acting aloneand sampling variability, we can assume that if the process was repeated, and/or a larger number to simulations were completed, our experimental frequency would be closer to the theoretical.

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In our simulation we visualised 184 runs of the chocolate milk promotion to see how many times our lucky numbers (7, 12 and 13) appeared in order to find the true probabilty of the chocolate milk. Compared to the 1 in 5 true probabilty our graph is reasonably similar to the true probabilty and although our results won't be the same, there's isn't enough solid visual evidence that our test runs do not reflect the true probability.

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We set up a simulation in order to gain a better understanding of the true probability of getting 2 out of 15 successes. This allowed us toy see trends in the probability and results, giving us an idea whether 2/15 was a likely rēsult. We did 184 test runs, enough to give a decent idea about the true probability of the different results. This showed that there is lots of variation due to random chance and both 2 and 4 were also very common. This means that getting 2 is not at all a strange occurrence and there is no reason to suspect fowl play.

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We simulated 184 random groups of 15 numbers each between 1-15 with selected numbers  being 7,12,13 were randomly selected and when these numbers were chosen during the simulation we would record how many times these numbers appeared during a set of 15 random numbers.

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We set up a simulation, where in each test we randomly selected 15 numbers in between 1-15. Because we believed this was the true probability. We had 3 “winning” numbers, 7, 12 and 13 which we’d then see how many times came up in each test of 15. We carried out 184 tests and found that the winning numbers coming up 2 times occurred the most in the 184 tests we carried out. This means that if the true probability is 1/5 or 3/15, so chance acting alone made 2 the most occurring number

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In the simulation had a sample of 15 numbers and got 184 runs and gather numbers (7, 12, 13) between numbers 1-15.1 in 5 should be a winning prize which means 3 should be the most popular number.we did a simulation to get the true probability of getting 3 out of 15. this allowed us to see results and see that 2 was more popular than the expected 3.

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We simulated 184 random sets of 15 numbers ranging between 1-15. In each set we recorded how many times 7,12&13 were randomly selected. 3 special numbers were chosen as we were
attempting to see if there was a 1 in 5 chance. In this simulation we found that it was actually more likely to have a 2 in 15 chance, however if this simulation was 1000 attempts we would expect to see a 1 in 5 chance.

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We used a simulation to compare the true probability of 1/5 with the experimental probability. We ran the simulation 184 times and found the most common probability was 2. This was not the expected, as 1/5 was the theoretical probability we expected 3/15 to be the most common. However due to randomness and sampling variability, 2 was the most common result.

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In our simulation, we had a sample size of 15 from 1-15. We choose our winning numbers were 7,12,13. We then ran the simulation 184 times. The probability we were testing for was 1/5 or 3/15 times that we get the winning numbers. Because of sampling variability, we aren't always going to get 3 winning numbers in our sample. From the graph, we can see that if we don't get 3 there is a high chance that we get either 2 or 4.

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We simulated 184 runs with the purpose of seeing the variability of sampling in the amount of prize winning bottles (successes) selected to see if 1.5 was the true probability and determine whether or not there was a problem

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In our simulation we did 184 runs (randomly selecting 15 numbers) to see if the true prop

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We took the statement that 1 in 5 bottles of chocolate milk contain a prize (the true probability). Naturally then 3 out of 15 will have a prize. We then chose the prize numbers - 7, 12, 13. Groups of 15 randomly generated numbers were then produced so we could see how many times the prize numbers came up. With this information we get the graph. Doing simulations allows us to visualise sampling variability. We get a better idea of the true probability (here presented as 1 in 5) by doing lots of trials in attempt to get a pattern. Here we see that getting '2 out of 15' came up the most (51 times) whereas gettig '3 out of 15' (the true probability) came up 47 times. However through chance acting alone it is plausible to see these results as 2 is pretty close to 3 and we may even see that 3 comes up to most with more trials (considering we only did 184 runs

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we did the simulation 184 times this is so we can see if if the true probability is 1 in 5, our runs had a size of 15 so in real life we could see if 3 out of 15 would work and to see the pattern. the three numbers we choose were 7,12 and 13 we generated this randomly. the graph shows me that the two was more common than 3 i would expect to see that this evens out after more tests. due to the numbers around 3 eg 2,4,1 all having quite high probability as well so this would not be out of the ordinary to see this happen due to chance acting alone on these unlike if it was 9 times as it is a lot further away.

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We set up a simulation where we simulated the true probability of getting a prize. We think the true probability is 1 in 5. In the simulation we have a sample of numbers from 1-15. We then used 7,12,13 as winning numbers and did 184 simulated runs. In each group we counted how many "successes" there were. In the simulation we found that 2 out of 15 was the most common (51) for how prizes were won per group. However, if we did 1000 simulated runs we might have seen 3 as a more common outcome. By chance acting alone we could still get 1 or 0 prizes as the percentage seen from the simulation is 16%.

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We simulated 184 runs. This allowed us to see the true probability r

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In our simulation, we randomly generated sets of 15 numbers from 1 to 15, and assigned 3 of those numbers to represent prize numbers, as the true probability of a winner in the scenario was 1 in 5. We then simulated these tests 184 times. 3 prize winners occurred 47 times out of 184, and 2 prize winners occurred 51 times. However, if we were to increase the simulation size it is likely that there would be more 3 prize winners, as 3 in 15 is the true probability. In any case, the fact that Alfred only won 2 prizes from fifteen bottles should not be too surprising, as the probability of this is around 51/184, or 27%, and was the most likely outcome in our situation. The reason that the outcome is not always 3 out of 15 as indicated by the true probability we assigned is due to sampling variability, as chance is acting alone.

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We set up a simulation where we found groups of 15 random numbers, as the person bought 15 bottles, and 2 of them were prize winning bottles, whereas we believe that the true probability of getting a prize winning bottle is 1 in 5. We did 184 runs and counted the number of successes there were, ie. how many sevens, twelves and thirteens there were. In our simulation, we got 8 zeros, 22 ones, 51 twos, and 47 threes. This shows us that every time you do a simulation, you will probably not get the same results again, because of sampling variability. Chance acting alone may mean you get less or more than 3 prize winning bottles for every 15 bottles you buy. As our simulation, there was one 9, which is far more than 3 in 15 prize winning bottles.

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Righto, after 184 tries you can see that there is a clear pattern with majority of the rounds of 15 bottle successes ending up with 1,2,3 and 4 times. If the true probability is 1/5 then this shows that with 184 tries you will see something similar to a 1/5 chance, you will not get exactly 1/5 all the time as the test is random. Based off this graph you will see that 2/15 was more frequent than 3/15, you would expect this as it is random. Some tries got 9/15 which is extraordinary but  due to the random nature of this test it does happen.