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. 😎

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Phillp Woodhouse Yay! I’ve got your back. 😎

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In which class do you study christopher?

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Zoom er Yay! That’s awesome. Glad I could help. 😊

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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. 😎

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It is 0 as their intersection is zero because of mutually exclusive charactiristic

mKut Man Yay! Thank you. 😊

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.

Gurr e Sekhon That will vary by teacher. I personally accept all forms unless I specify otherwise in the directions.

I made a video for you: kzfaq.info/get/bejne/mNWpmppersnXdn0.html

(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.

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Thanks!

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.

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Ron Yay! I got your back. 😎

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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.

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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!!!

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P(A/B)=p(A and B)÷p(B) 2400/4000 ÷1600/4000

I know. I wish you were in my class! I could use another self-motivated student. :-)

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.

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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.

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You just just made my day. Thank you so much!

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 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): kzfaq.info/get/bejne/ZryhoM6h15fLonU.html

MrHelpfulNotHurtful thank you so much for your detailed reply! That makes sense

Solly Rella You are very welcome. 😊

That sounds so frustrating. I'm sorry you have to go through this.

glad I could help

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. :-)

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.

LOL. I will have to check him out.

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: kzfaq.info/sun/PLUq8yM4tK_aXJQkVr6d-84R9AaAuwWYt2 I did a few things differently, so check it out. You might like some of my new strategies better. :-)

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@nut so you've chosen to be a shitty human, huh!

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