Hello, in the video about independent events, Greg mentioned that when we pulled one red marble, the probability of choosing a red one is 5/10. Then with no replacement, the probability of choosing a red marble would be 4/9. So, as you can see, with no replacement, the probability decreased.
On the other hand, in the first video about probability introduction, it was mentioned that regardless of whether we put things back inside the bag or keep them out, the probability of fifth or tenth would always be the same as that of the first. And this concept was even emphasized in a question in the related quiz.
The issue is, I find these two things contradictory. In the first case, the probability went down when we didn’t put the marble back; whereas in the second case, the probability remained the same with no replacement.
Could you please elaborate on how we can reconcile these two ideas?
What you should notice is that there are as many arrangements where there are red marbles in the third position vs where it’s in the 10th position. This means that the probability in Quantity A should be the the same as Quantity B because:
Required probability = (all arrangements that you require)/ (total arrangements)
and the “arrangements that you require” happens to be the same for QA and QB for the aforementioned reasons.
More rigorously, this means there’s a self inverse bijection, and so the cardinality and hence the probability (uniformly distributed) is equal.
In my original question, I was looking at 2 cases that looked contradictory. The first case was pulling out marbles one by one without replacement; and the second case was how many times a specific marble was chosen first or third or fourth. I figured out that I thought of the second case as separate events, which was why both cases seemed contradictory to me.
Based on your other question, I have a feeling that you’re conflating two unrelated things. This question has little to do with independence as you might think it does. Sure, there are two events, but events are just subsets of the sample space, so I’m not sure what your intent is post identifying two such events.
I think you should try to understand my explanation above, as that is the only right way of going about this question.
To reiterate, there are equal number of arrangements where red marbles occur in the third or the tenth position. You also have that every single arrangement of the 50 marbles have equal probability. Putting these two together, you can then conclude that Quantity A = Quantity B.
I got your point here. But what makes it hard for me to absorb this concept is the following:
-I agree that the probabilities are the same for each arrangement (third or tenth)
-However, the “with no replacement” part makes me deduce that the total number of elements, after throwing out let’s say the marble that came third in pull 1, would now be less by 1. So, the total number of marbles right before pull 2 would be 49, thereby altering the probability of an arrangement with a marble coming tenth. Both the numerator and denominator would change due to no replacement after pull 1.
Pulling red marble third is the same as pulling 10 marbles and looking only at the 3rd element.
_ _ R is the same as _ _ R _ _ _ _ _ _ _
You can compare this with pulling red marble on the tenth position:
_ _ _ _ _ _ _ _ _ R
The question assumes you pull all 50 marbles, but even if you stop after the 3rd pull in QA and the 10th pull in QB, the answer remains the same.
If you still don’t get it then I think you should just skip the question. I’m sure it’s not that important, and all this time could be spent on some other topics. Understanding bijections present in a question like this enables you to solve combinatorics problems of this form (another question someone else posted from gregmat):
If the digits 7,7,3,2, and 1 are randomly arranged from left to right, what is the probability that both 7s are to the left of 1?
With the same idea, you know that exactly \frac 13 of the arrangements have both 7s to the left of 1. Did you need to identify that for the question? No, because the casework was scalable in this case.
Thanks a lot Cylverxxx for your explanations and examples. They are much appreciated. For the question in the photo you attached, I wrote down my train of thoughts. Could you please review it and let me know if I am doing anything wrong?
I understand that the probability of 771 coming first or second or third is always the same. But the new question that arose this time is, should we multiply the 1/3 by 6, since we have 6 arrangements?
No, i guess you’ve done an entirely different problem. First of all, you can’t treat 771 as one object because there are many valid arrangements you’re skipping over.
Some arrangements that you totally gloss over would be:
77231
72713
…
The question just states that both 7s have to be to the left or 1 and nothing about them being together.
Why don’t you try the “mainstream” way of solving this question using basic combinatorics? Like if you saw this question on your exam, how would you approach it?
Hint
Do casework by fixing 1 in the first position, second position, and so on.
I could always just give you the answer, but it’s more instructive for you to try and keep getting stuck, since that serves as a learning experience.
No, you didn’t take my advice with the casework. The 7’s don’t have to be together either so something like 7 3 7 1 2 also works.
Anyway, building off what i previously suggested you to do:
Case 1: 1 _ _ _ _ _ → No 7’s can be to the left of 1
Case 2: _ 1 _ _ _ → Both 7s can’t be to the left of 1
Case 3: _ _ 1 _ _ → Both 7s have to before 1 for this case to work. This leaves you with 2 \& 3 occupying the 4th and 5th slot. There’s 2! ways for such arrangements.
Case 4: _ _ _ 1 _ → …
Case 5: _ _ _ _ 1 → …
I’ll leave as exercise for you to fill out the last two cases.
with the casework. The 7’s don’t have to be together either so something like 7 3 7 1 2 also works.