Y1 Probability Distribution Q

A fair 5 sided spinner is spun.
Given the spinner is spun 5 times, write down, in table form, the probability distributions of the following variables:
Red - 2/5, Yellow - 1/5, Blue - 1/5
A) X, the number of times red appears.

I did this as
x = 0, (3/5)^5
x= 1, (2/5)(3/5)^4 etc. etc

But the mark scheme shows this combined with the values from pascals triangle i.e. like a binomial expansion - so (2/5)^5 for x = 0, 5 * (2/5)(3/5)^4 for x = 1.

I just don't understand why this is done. I get that it technically fills the conditions, but there's a question used in the example for this chapter (binomial distributions is the next chapter) where it's an identical question but they don't use binomial distribution?

The question is:

This spinner (red - 2/5 , blue - 3/5) spins until it lands on red or has been spun four times. Find the probability distribution of the random variable S, the number of times the spinner is spun. Here, they do x = 1 is 2/5 , x = 2 is (2/5 * 3/5) etc etc.

Is it because the first question is spun 5 times for definite and the second isn't? Thinking that must be it, but confused as to why that question is in this chapter when we aren't introduced to binomial distributions in stats until the next one...

I think they used the binomial distribution because they regarded P(successfully landing on red) and q(not landing on red)=1-p. I guess they used binomials to save time, because if you were to draw a tree diagram or a sample space, it takes time

In the second example, it’s geometric distribution which is when you keep trying until you reach your first success, that’s why it’s differently done, it has P(X=r)=pq^r-1

I hope this helped a little

I don't understand the question - can you copy it exactly? Is line 3 giving the probability of each colour occurring per spin? But then they don't sum to 1.

Assuming there is a 2/5 chance of getting red on a spin, I can see a mistake for your X=1. Think about how many different ways you could get 1 red and 4 other colours. Hint: it isn't 1.

I've never heard of it being described as geometric distribution! Will have to google the P9X=r) pq^r-1 thing as I'm struggling to see how the r is in that...

I think it’s a new topic, it was just recently introduced in the board I’m following (CIE), r represents the first success, so for example if takes 4 spins until you land on red, then r=4

ah yeah - that makes sense - so general form for the process of multiplying "failures" by themself until you hit your success. So if you take 1 spin to land on red, then P(X=x) is (2/5) as it's pq^1-1

Yes!! You’re getting it right, just in case you run into questions where they require you to use it, these are formulae I found useful

I see how the red could be achieved at different times, but not necessarily in different ways?

Different ways / different times = same thing. So do you see how many different ways you could get one red spin?

5?

Yup that's right. So your (2/5)(3/5)^4 is correct, but only for one of those ways. Since there are five overall, you add them all together i.e. 5*(2/5)(3/5)^4.

Oh - so P(1 red) = P(1 red, 4 other) AND P(1 other, 1 red, 3 other) AND .... so basically 5*(2/5)(3/5)^4 since on a tree diagram, for an example, all those possibilities would sprout off each other, so you have to multiply across?

Add - because, again, if it was a tree diagram, I'm looking at the probabilities of all those combinations and adding them together. The probability of those combinations all happen to be equal, and there are 5 combinations, so I can just multiply by 5. Right?