Do we need to know this?

[True Dat]

Hi everyone,
Was just wondering if we need to know how to do this sort of method for the methods exams coming up tomorrow.
I've attached worked solutions of the kind's of questions that involve the theory that I'm wondering if we have to know, as it never showed up in my method's 3/4 textbooks ever.
Thanks

[Quantum44]

Hi everyone,
Was just wondering if we need to know how to do this sort of method for the methods exams coming up tomorrow.
I've attached worked solutions of the kind's of questions that involve the theory that I'm wondering if we have to know, as it never showed up in my method's 3/4 textbooks ever.
Thanks

if they give you a small population, using combinations to work out the distribution is the easiest method as binomial can only be used on large populations since technically when you choose a person for a sample, you reduce the size of the population you can choose the next person for your sample from. I'm fairly sure this is covered in the Cambridge 3/4 textbook.

[True Dat]

if they give you a small population, using combinations to work out the distribution is the easiest method as binomial can only be used on large populations since technically when you choose a person for a sample, you reduce the size of the population you can choose the next person for your sample from. I'm fairly sure this is covered in the Cambridge 3/4 textbook.

I remember combinations and permutations in Unit 1/2 but so far in  my Nelson textbook, there hasn't been any mention of them, so is it just presumed knowledge or?

[VanillaRice]

I remember combinations and permutations in Unit 1/2 but so far in  my Nelson textbook, there hasn't been any mention of them, so is it just presumed knowledge or?

Not sure about your textbook, but it's definitely covered in the Unit 3/4 Cambridge textbook. There should be something in your textbook under sampling (especially sampling from a small population)? I'd be pretty surprised if it wasn't covered. If not, have a search of the "hypergeometric distribution". You only need to know how to do calculations like the one done in the solutions you've attached. You don't need to know/learn anything else about the hypergeometric distribution (mean, SD, etc.)

[atar.notes.user]

when exactly do we use combinations? like what 'word' in the question will give us the hint to use it?

[keltingmeith]

I'm gonna dispel a couple of myths, because I thought it seemed weird that this was asked. I double-checked the study design, and I can promise you - you DON'T need to know how to do this. The study design does NOT talk about using combinations or permutations in 3/4 methods AT ALL. They do point out you may be expected to do questions like this, but very explicitly specify only in cases where the ARITHMETIC is simple - so, just +, -, x, and /.

when exactly do we use combinations? like what 'word' in the question will give us the hint to use it?

This is a very flawed way of thinking - do NOT try to answer questions just by looking for a particular word, it'll save you a world of hurt.

[Quantum44]

when exactly do we use combinations? like what 'word' in the question will give us the hint to use it?

If there's a stats question with a small population. For example if you are calculating probabilities of p hat when taking samples of 4 students from a class of 6 boys and 9 girls for the proportion of boys in the sample. P hat can be 0, 1/4, 1/2, 3/4 or 1 but you can use combinations to create a probability distribution.

[Alwin]

I don't meant to turn into TrueTears and give you a full blow-by-blow account of probability and statistics, but to my eye that looks like a Hypergeometric distribution which is used for selection without replacement. While you don't need to know what a hypergeometric sequence is (esp. so close to the exam!) but it's based on combinatorics as others have stated. You can think of it as a quicker alternative to drawing out a tree diagram.

Essentially:
Take a sample of 2 from a group of 8. You can't expect the probability to remain the same since the probabilities will significantly change between selections! You can draw out a tree diagram and you'll find:

You'll notice when we sum these two together you get to the same probability as the worked solution,

A binomial distribution could not have been used because the probability of success/failure changes for each 'step' in the tree diagram.

For a larger population when you select a few cars for a sample you can do the same thing of changing the probabilities as each item is selected and drag a tree diagram etc. On the other hand for very large samples, e.g. 1000, then 500/1000 is approximately 499/900. So we can make the assumption that the probability is constant even though selecting a sample is technically a non-replacement process meaning we can use a binomial distribution.

However, typically in Methods it's a clear cut case where they tell you the population probability is P (this is almost always the case for statistics!!).
It's only when they tell you the population size is Y and a sample of size of X where X and Y are similar magnitude you need to think about creating a tree diagram/using combinatorics.