Day 7 HW Conditional Probability + Independent vs Dependent Events
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- Опубликовано: 12 янв 2015
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Here is another more recent probability playlist of mine:
• Geometry - Probability...
Thank you so much! So direct and straight to the point! I love you for this!!
Yay!! So glad I could help. :-)
Thank you ! I have been struggling with these type of problems for 3 days.
You are very welcome!
"target over total"..been struggling and as soon as you said that it made sense. and those tests of independence, thanks ! i get it. finally
You are very welcome! So glad I could help. :-)
Great review before my quiz thanks!
So glad I could help you. :-)
Thank you. In just the first two minutes, you have given me the answers I have been searching for.
Phillp Woodhouse Yay! I’ve got your back. 😎
Very straightforward presentation. Thank you!
You are welcome, my friend!
So exited to have this. You make teaching very practical and real. Thanks
In which class do you study christopher?
thank you , you're the only one who made it clear that p(A and B) = p(A) . p(B) only when they are independent
You are super welcome!
thank you for making this video! very helpful:))
So glad I could help. :-)
This is so helpful! I really needed something to help me understand so I can help my daughter with her homework. Thanks!!
Helping your kids with their math is tough! I’m glad I could help. I have full courses worth of videos through calculus. Use my channel as a resource. 😎
Thank you so much for your time!
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Extremely very helpful explanation!!!....thanks a lot!!!!
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thank you so much! watching this as my final revision for my exam tomorrow
You are super welcome. Good luck on your exam! 🍀
Thank you so much. For sure I will pass the exam!
You are super welcome! Glad I could help.
So helpful! Thanks for sharing this.
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Thanks so much for this. I now know so much more about the 7hw lore.
😄
excellent, maths weigh over my head is now solve by your step by step system
Thanks! I'm glad you are finding my videos helpful. :-)
Thank you so much the best explanation ever !
Yay! You are very very welcome!
Your illustrations have been clear and hence made my studying easy for my exams. Thank you all the way form Kenya
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Thanks a lot! I figure out the answer to the question I was stuck on with the first minute of your vid.
Zoom er Yay! That’s awesome. Glad I could help. 😊
extremely helpful, will definitely subscribe!
So glad I could help. :-)
Great for my exams preparation !!
Awesome! So glad I could help.
fantastic. awesome explanation. best teacher ever. thanks a lot :)
Aw. That makes my day! You are so very welcome. :-)
Very helpful! Thanks
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May God bless you .
Thank you so much. :-)
Excellent,well understood
Awesome. Glad I could help. :-)
Thank you alot for your explanation, it's helped me
You are very welcome my dude.
Thank you so much! This is super good :)
Hello Janes Yay! Glad I could help. 😎
Thanks for your video
You are very welcome, my friend.
Great!!! Blessings from Nigeria.
Bunmi Chukwu Yay! Glad I could help. 😊
V clear and made simple,thanks
You are quite welcome. :-)
I hate how I found these after my school semester finished. The time where I don’t wanna study lol… AND THEN I GET THESE! Such good ones alos
I wish you had found them sooner! Look around the channel to see what else is there. I have playlists by subject and unit. Find your next unit and watch the videos ahead of time. 😎
Genius! Made it simple.
Yay! So glad i could help. :-)
Thank you🤗
🥰
well done thank you
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Thank you!!!
You are so very welcome. 😊
very clear explenation
Thanks!
thanks!!!!!! this was so helpful
Yay! So glad I could help. :-)
You so Good...Respect!!
mKut Man Yay! Thank you. 😊
very helpful and i have learnt
Yay. I've got your back.
Wish I could do my homework with you!
I know. I wish you were in my class! I could use another self-motivated student. :-)
My teacher is horrible at explaining and teaching us, and it is especially more frustrating when I go and ask him for help. It is like he can give me the answer, or show the process to me and hooe I understand it, or I endlessly guess on how to do something I have no clue what so do.
Good one.
Thanks!
great indeed
Yay. 😊
You ma friend are amazing. you made ne have an Aha!!! moment. thanks
You just just made my day. Thank you so much!
Thanks !
You are super welcome :-)
Is it necessary to put answers always in percentage. Does it effect marks if we put them in decimels. Thanks
Gurr e Sekhon That will vary by teacher. I personally accept all forms unless I specify otherwise in the directions.
Question. Why are you dividing by 100 and 8, etc. on problem 2?
(I think you meant problem 3 so I will answer based on that.) Probability can be written in three ways: fraction, decimal and percent. In this video I tried to show the answer in all three forms. For example: 2/3 = .6667 = 66.67%. If you want your answer as a decimal or a percent, you don't have to divide by 100, 8, etc. You can just divide the two numbers you have like 1600/2400 = .6667 = 66.67%. However, if you want to see the probability as a fraction, you need to reduce the fraction. You know for example that 6/10 reduces to 3/5; I divided the numerator and denominator by 2. Well, 1600/2400 can be reduced in a similar way. The 00 and the end of the numerator and denominator tells me that they are both divisible by 100. If you divide both by 100, you get 16/24 (you basically cross out the zeros). Again, I am just reducing the fraction. Now I need to reduce 16/24. Both numbers are divisible by 2, they are also divisible by 4 and 8. It is quickest to divide by the biggest number you can, so I choose to divide the numerator and denominator by 8. This gives me 2/3.This is the faction form of the probability we were looking for. I just divided by 100 and then 8 in order to reduce the fraction 1600/2400 down to 2/3.
Thank you so much for explaining with such detail. I sincerely appreciate it. From someone who is trying to become friends with statistics... Urgh! But anyhow, I think I just mastered the "Given that" problems with your help. Hahaha! Thank you again! ;) I look forward to your videos to get me through this semester.
Made it so easy!
Ron Yay! I got your back. 😎
how could i solve in case of mutually exclusive which probabilities are P(A); (0
It is 0 as their intersection is zero because of mutually exclusive charactiristic
That was verygood
Thanks! :-)
please give ansewer for this question a manager wants to assign 20 workers to four distinct construction job these jobs require 6 4 3 and 7 workers respectively. in how many different ways can the manager assign the workers?
I made a video for you: ruclips.net/video/boyg7-IcsFI/видео.html
hi, thank you for the video. in 18:39 (problem 6) I am not sure how are these events are dependent on each other? they seem pretty independent to me. Same goes to problem 5, how are these events independent if they share the outcome "3"... Sorry I am struggling with understanding this but I do understand the calculation.
Faisal Alfaiz I think you are confusing the concept of dependent vs. independent with overlapping vs. mutually exclusive. When you are trying to judge whether events are independent, you are not meant to use your intuition to figure it out. The definition is based on the calculation. You are just meant to do the calculation and trust the result.
Thanks
Solly Rella You are very welcome. 😊
Hey man Honestly im really confused about probability can you make a video of explaining the rules of the conditional prob
I don't actually know a set of rules for conditional probability. When I see a problem, I can figure out how to solve it and I can explain how I did it, but for conditional probability, I don't think in terms of rules.
wouldn't the probability of A and B should be 1/5? because A has 3 events (1,3,5) and B has 2 (3,6). If not, can you explain why is it 1/6?
Event A = rolling an odd number
Event B = rolling a 3 or a 6
Therefore Event A AND B = (rolling an odd number) AND (rolling 3 or 6) at the same time.
The only way to roll an odd number AND roll a 3 or 6 is to roll a 3.
Since there is only 1 winning value out of a total of 6 numbers on the cube, your chances are 1 out of 6 or 1/6.
For question 1, is it should be a conditional probability problem? Let's define A=winner win the gold medal and B=winner from the United States, my understanding is that the question is asking P(A|B).
P(A)=301/928=0.324; P(B)=97/928=0.105; P(A and B)=39/928=0.042. Therefore, P(A|B)=P(A and B)/P(B)=0.4.
Please correct me, thank you!
Notice that your answer (0.4) and my answer (0.4021) are very close. The bottom line is that your answer is a tiny bit off because you are rounding and I am not. You say that P(B)=97/928 = 0.105, but it does not because 97/928=0.104525862… You say that P(A and B)=39/928=0.042 but it does not because 39/928=0.042025862… Let’s do the problem using your method, but without rounding.
P(A|B)=P(A and B)/P(B)
P(A|B)=(39/928)/(97/928)
P(A|B)=(39/928)*(928/97)
P(A|B)=(39/97)
P(A|B)=0.402061856…
P(A|B) is about 0.4021 or 40.21%
I think the method that I used in the video is a lot easier though.
That makes sense! Thank you!!!
You are very welcome my friend. :-)
P(A/B)=p(A and B)÷p(B)
2400/4000 ÷1600/4000
thanks
glad I could help
is there any other formula that applies for independent and doesn't apply for the dependent?
To be honest, I haven't taught this for a few years so I don't remember the details anymore. Sorry :-(
never mind , you're still the only one who explained conditional and dependent and independent very clear
Can you help me please
1-A company has three machines B1, B2, and B3 for making 1 k resistors. It has
been observed that 80% of resistors produced by B1 are within 50 of the nominal
value. Machine B2 produces 90% of resistors within 50 of the nominal value. The
percentage for machine B3 is 60%. Each hour, machine B1 produces 3000 resistors,
B2 produces 4000 resistors, and B3 produces 3000 resistors. All of the resistors are
mixed together at random in one bin and packed for shipment. What is the probability
that the company ships a resistor that is within 50 of the nominal value?
2-Suppose that we have two bags each containing black and white balls. One bag contains three times as many white balls as blacks. The other bag contains three times as many black balls as white. Suppose we choose one of these bags at random. For this bag we select five balls at random, replacing each ball after it has been selected. The result is that we find 4 white balls and one black. What is the probability that we were using the bag with mainly white balls?
Sherouk Taresh This is a bit over my head I’m afraid. What class is this for? I teach geometry to HS sophomores. We don’t go very hard core.
I was following till 15:40
I just re-watched it to refresh my memory. If you can think of a specific question you have, I may be able to help. It is really tough to explain math in a comment. I fear that a general explanation here would be more confusing than the video was. However, something very specific is often doable. :-)
I have a conceptual problem with the independent vs. dependent examples: Two consecutive rolls of a fair die (or a single roll of two fair dice) should be independent no matter what. However, whether the result of the second roll can satisfy both conditions is dependent on the conditions stated in the first event. I'm not sure how to reconcile these two views.
Can you give me a problem or two quoted from a textbook or at least a teacher illustrating the contradiction you are trying to reconcile. When it comes to probability, it's all about the details. Your initial question is too vague. Let's nail it down with some concrete examples.
Sure, I was referring to 12:00 and onwards in this video. So for instance, the intersection between 5. These cannot exist simultaneously, so the question poses two disjoint outcomes, which are not independent. However, the dice rolls that produced them are independent events, since the outcome of one roll has no effect on the outcome of the second roll.
Here is where I believe your mistake is. You are saying that 5 cannot happen simultaneously. However, you go on to explain that there are two dice (or two rolls). If there are two dice, one could be 5. Therefore the two events CAN exist simultaneously. The two events are independent.
On the other hand, imagine that you only have one die and you only get one roll. Event A is rolling 5. Since we only have one die and one roll, these events are mutually exclusive (they cannot both happen). In this case Event A and Event B are not independent (AKA dependent).
Even though there is a connection, do not make the mistake of simply thinking mutually exclusive = dependent, and not mutually exclusive = independent. They are two separate concepts that should be thought of individually. Here is a helpful video (not mine): ruclips.net/video/0Vqmkpr1grA/видео.html
MrHelpfulNotHurtful thank you so much for your detailed reply! That makes sense
What if you already know what p(a) and p(b) is but you have to find p(a|b)
Give me the details of the problem and i will work it out for you. If a chart is involved, take a picture of it, store it in the cloud and post a link to it here.
I hate that nothing really helps me unless someone is helping me one one. It I don't understand something that is most likely the only way it will help, and I tried tutoring it doesn't help me as well. It must be the exact style of problems I am Doing, and go though every variation for me to understand it pretty much.
That sounds so frustrating. I'm sorry you have to go through this.
This can't be right. (@ 1:30) you state the P(A and B) = P(A)*P(B) and you also state that P(A|B)=P(A and B)/P(B) which is the same as P(A|B) = (P(A)*P(B))/P(B) the P(B)'s cancel and you get P(A|B) = P(A); this is not correct.
The first formula is for independent events. The second formula is for conditional probability. You can't substitute one formula into the other because they are each used under different circumstances. If you give me an actual problem, I can explain which formula we use and how we use it.
@@MrHelpfulNotHurtful you should make this distinction in your video. Otherwise point states valid.
You kinda sound like Chance the rapper.
LOL. I will have to check him out.
Sooo hurtful
Sowwy
MrHelpfulNotHurtful jk haha, i watched almost every probability video u uploaded. i was kinda confused but i think its easy now...
i still have two chapters to study, i hope u have it in ur playlist...
@@ALbaraa2X Here is a whole other probability playlist that I made recently: ruclips.net/p/PLUq8yM4tK_aXJQkVr6d-84R9AaAuwWYt2
I did a few things differently, so check it out. You might like some of my new strategies better. :-)
When you copy an entire video from khan academy.
?
@nut so you've chosen to be a shitty human, huh!
thanks
You are super welcome