Statistics 101: Combinations - Losing Your Marbles
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- เผยแพร่เมื่อ 15 ต.ค. 2024
- Statistics 101: Combinations - Losing Your Marbles.
In this video, we cover, step by step, one of the basic types of combination problems; selecting groups based on conditional criteria. If you are in an intro stats or finite math course you WILL see this type of problem. I walk you through step by step.
If you are having problems in a course, please stay positive and keep your head up. To be studying this you have to have accomplished quite a lot already. Have faith in yourself and keep learning.
My playlist table of contents, Video Companion Guide PDF documents, and file downloads can be found on my website: www.bcfoltz.com
Brandon, you are a lifesaver. I've really been enjoying your videos over the past several weeks as I'm working my way through an online statistics course. Your videos are an excellent supplement to the course material (actually, better than the course material). I also appreciate your messages at the beginning of each video that cheer me on and tell me to stay positive; it really does help. Thank you, thank you, thank you.
really appreciate how encouraging you are and how concrete you make it
Wondrous explanations! BF sets the benchmark for excellence in teaching difficult topics. Very well done!
This really helped me. I’m just returning to college after 10 years away. It’s not that easy, but this video helped. Thank you.
i'm literally losing my marbles because of math but u saved me!!! ur such a lifesaver
I'm studying for the DAT and these exact types of problems are going to be on the quantitative reasoning section.
Like you said I think in a very linear fashion and I've been having trouble finding good explanations on solving these problems but your videos have made it clear on the reasoning. I'm going to work on more problems and get it solid in my head now. THANK YOU!!!!!!!
wow!!! I was struggling with the concept for a very long time... but now its crystal clear.... thanx Bandon
You are an excellent teacher, keep up the good work , thank you so much sir
Awesome! Thank you so much. Keep working hard. I know you will do great! - B
Thank you very helpful! I almost lost hope but then I came across the video.
Thank you! This video really helped me understand combinations with restrictions and the positive message was nice too.
Very good explanation together with good examples.
Thanks
Thank you so much for all these wonderful videos, Brandon. You are such a wonderful and encouraging teacher! I don't know if you still check on the comments on this course but I am going to give it a go. Here is my question: If I take a sample of 100 out of a population of 10,000 marbles, which consists of blue, red and green, and manually count the number of marbles in different colours. What is the best way to calculate the probabilities of the marbles in blue, red and green in the whole population? Many thanks in advance.
Hello! There is a test for population proportions. This video _might_ help: th-cam.com/video/SbYv6fYtTG0/w-d-xo.html The question is do you _know_ the proportion of colors in the 10,000 or are you just estimating is based on the sample?
@@BrandonFoltz Wow Thanks very much for getting back to me, Brandon. In terms of the question, it is purely based on the sample proportion to make an estimation of the whole population distribution of the colours. Would that be possible?
Hey Brandon. I love your videos.
I have one confusion. For the 3rd question, where atleast 2 are red.
Since it's atleast, it doesn't really matter if the 3rd and 4th are red or not. So what if we simply calculate it as :
C(4,2)*C(12,2)= 6*66= 396.
Similarly for 4th question, what if we calculate it as :
C(4,1)*C(9,3)= 4*84= 336.
Please help.
I have the ssame doubt, except for the 'at least 2 red' case I thought it wasss C(4,2)*C(14,2) --since after selecting the 2 reds, we would have 16-2 = 14 balls left to make the other 2 choices. Similarly, for the 4th question, I thought it was C(4,1)*C(15,3)
@@capelsid There are only 14 balls in total
@Mugdha, C(4,2)*C(12,2) will lead to double counting of some combinations. Some among the six combinations that come out from C(4,2) and the combinations that come out of C(12,2) will be the same. So, if we multiply the nos. of resultant combinations arising out of these two (C(4,2)*C(2,2)), we will not be eliminating the repeated combinations.
Awesome lecture - thanks !
this was so helpful thank you!!!!!
Great explanations! Great job.
Though I understood the concept...the last bit ( Reverse way of solving) is a bit challenging...But as you said, 2-3 times of watching this video will give me the insight...Thank you Brandon !!!
very impressive teaching
Excellent...
Awesome video - thanks!
Unlike the title, I got all my marbles back :P Thank you!!
Escellent... Thank you so much.
Hello, sir, thank you very much for your help! could clear why do we consider marbles of the same color different? why R1R2BG and R2R1BG are not the same combination?
"Each marble is labeled with a number so thay can be distinguished. "
It's written in the problem. Play the video in 2:37.
In the 2 of 4 red, 2 not red out of 10... shouldn't this be 2 not red out of 8 (because we have 4 red total, so in the other 10 left in the bag we still can pick 2 red and this will make more that 2 red in the group we pick ) ?
I thought that once you grab marbles they are not available to form another sets.
but what if I want at least 2 red and 1 green?
God bless u.. thanx
Thank you!
Is this the addition rule?
where did you get 1001 possibilities?😕
7:11
please teach my class. My teacher sucks at explaining this!!
Third problem can also be solved in "Reverse Way".
probably should be in 25:00, all possible with no red AND green (only yellow and blue)
haha if i was right.. I think I learned this on my 7th grade?
Way Way better than any MOOC