I love this guy. He is brilliant at explaining things. I love the simplicity of his approach. Just a piece of paper, a pen and a go pro and his own incredible talent explaining things simply..
I was confuse in numerator and denominators but here is one one trick you must always to see the random variable value i mean how much ball your are going to choose, out of 3 blue and 5 Red , as random variable is 3 for blue so we will begin from zero to 3
Use the calculator function "NCR", to calculate the numerators and dominators. so enter (x NCR button on calculator then the small number. so for the first example 0: its "3 'NCRbutton "0" * "5 NCRbutton 2" this gives you the answer 10. Now I want 2hrs of my life back looking for this
Aya Elayyan I'm not sure how to simplify it for you in writing but i'l try with the help of God... FOR: C0= You are instructed to choose THREE BALLS AT RANDOM (remember: X= No. of blue balls) , and again you are told x=0 [[which means ZERO BLUE BALLS... Given that you know 3=balls are blue AND 5=balls are red]] You have to simply choose ZERO FROM THE 3 BLUE BALLS and THREE FROM THE 5 RED BALLS as you are INSTRUCTED TO CHOOSE THREE BALLS AT RANDOM (3C0*5C3/8C3) Divided by the total!! that you choose---- dO the same for C1 and C2 however should u require more explanation i dont mind IF I HELPED U
firstly thank you soo much ^^ Smanga Sanga but this is not what i didn't understand :P i want to know how he found that 8C3(the total) Equals 56 for example and the same thing for 3C0 and 8C3 ..How he found the numbers ?
Aya Elayyan I apologize for answering the wrong question... I figured that the TOTAL is 8 which is 3=blue balls PLUS 5=red balls... and from that you choose 3 balls at random... Because that is what you have to choose at the end of the day according to the question e.g. regardless of you choosing 0=blue balls (3C0*5C3)....@the end you have to choose THREE BALLS. Hence it is 8C3 which is the denominator part (total)
5 choose 3 divided by 8 choose 3 Since you have 5 and you want to choose 3 from them to not get a blue, you say "five choose three" as in "out of five, choose 3 of them". And the same goes for 8 choose 3, except the 8 is the total possibilities you could have chosen (ie. the three blue + five red), but we are still choosing three. The "X choose Y" stuff refers to the Combination formula.
I have a very quick question about finding P(x=1). Why can't you do the multiplication rule to find P(x=1)? I thought doing (3/8) * (5/7) * (4/6) would give you P(x=1). Why wouldn't it work for this case?
+brandong Since we do not care about the order of the selection, we are better off using combinations. Looking at your tentative solution, you are implicitly considering an ordered selection: your calculation is really saying a blue ball first, then a red ball and finally another red ball (let us call this b,r,r). BUT, you can also have r,b,r and r,r,b. You can see from this that in an ordered setup, P(X=1) = P(b,r,r)+P(r,b,r)+P(r,r,b) = (3/8)*(5/7)*(4/6)+(5/8)*(3/7)*(4/6)+(5/8)*(4/7)*(3/6), which yields the same answer.
Why do we need to use nCr when we have 3 slots for red/blue balls? Why don't we just have 8 combinations such as: RRR, RRB, RBR, RBB, BBB, BBR, BRB, BRR? But if order doesn't matter, then we should have: RRR, RRB, RBB, BBB 8C3 would give us combinations that repeat such as RRR.
What if there were 30 blue and 50 red balls? Then your first suggestion becomes completely unmanageable. To understand why combinations must be used, imagine numbering the balls in this case from 1 to 8: say 1,2,3 are blue and 4,5,6,7,8 are red. Hope this helps.
rose Decatur Hi Rose, it is a classic result of combinatorics that the number of ways of selecting (unordered selection and without replacement) r elements from a set that contains n elements is given by nCr = (n factorial)/( (n-r)factorial * (r)factorial). The expression (n)factorial, usually denoted by (n!), stands for the product of the first n positive integers, so for example (5)factorial = 5! = 1*2*3*4*5=120. I hope this helps!
I suggest you watch the following two videos on permutations and combinations: ruclips.net/video/vuIKbjoKegg/видео.html and ruclips.net/video/Z0rezieX1G4/видео.html. It should clear everything up.
I wish I knew the calculator commands >.< It's making me super lost. I might have known them a few years ago but I've been out of math for 2 years now and so my TI-84 is like a maze
+الفارس الأسمر X=0 means that we are choosing zero blue balls. Since we are choosing 3 balls in total, we are therefore choosing 0 balls from the 3 blue balls and 3 balls from the 5 red balls, hence (3C0)*(5C3).
You do need a basic understanding of combinations. You can check out my web page for statistics: web2.slc.qc.ca/pcamire/201-510-LW.html and watch the videos under section 5.
What you need is basic knowledge of permutations and combinations. I suggest you visit my web page: web2.slc.qc.ca/pcamire/201-510-LW.html and watch the videos of section 5. Afterwards, things will make a lot more sense.
I love this guy. He is brilliant at explaining things. I love the simplicity of his approach. Just a piece of paper, a pen and a go pro and his own incredible talent explaining things simply..
This was an excellent lesson I understood more in those 20 minutes than I did in 1 week of classes
You're way better than my stats prof.
man, your explanations are simple and accurate. Thank you so much.
Thank you so so much!! I was preparing for my probability theory exam and your video really helped me!!
I was confuse in numerator and denominators but here is one one trick you must always to see the random variable value i mean how much ball your are going to choose, out of 3 blue and 5 Red , as random variable is 3 for blue so we will begin from zero to 3
Use the calculator function "NCR", to calculate the numerators and dominators. so enter (x NCR button on calculator then the small number. so for the first example 0: its "3 'NCRbutton "0" * "5 NCRbutton 2" this gives you the answer 10. Now I want 2hrs of my life back looking for this
your math skills are on fleek
THANKS A LOT!! it was stuck on this for like an hour and every video made it even more confusing! Thanks!!! finals tomorrow so this really helped!
Very good Lecture....Specially your way of explanation.
Very good explanation... Use of combinations concept makes the probability calculations easier.
Really liked the concept of using combinations in calculating the probabilities.
Combinations are remarkably useful!
Very good explaination... Thanks you've assisted me tremendously.
but how he found the values of probabilities i didn't get it !
what's c2 and c0 and c1 equals to plz ?!
Aya Elayyan I'm not sure how to simplify it for you in writing but i'l try with the help of God... FOR: C0= You are instructed to choose THREE BALLS AT RANDOM (remember: X= No. of blue balls) , and again you are told x=0 [[which means ZERO BLUE BALLS... Given that you know 3=balls are blue AND 5=balls are red]] You have to simply choose ZERO FROM THE 3 BLUE BALLS and THREE FROM THE 5 RED BALLS as you are INSTRUCTED TO CHOOSE THREE BALLS AT RANDOM (3C0*5C3/8C3) Divided by the total!! that you choose---- dO the same for C1 and C2 however should u require more explanation i dont mind IF I HELPED U
firstly thank you soo much ^^ Smanga Sanga
but this is not what i didn't understand :P
i want to know how he found that 8C3(the total) Equals 56 for example and the same thing for 3C0 and 8C3 ..How he found the numbers ?
Aya Elayyan I apologize for answering the wrong question... I figured that the TOTAL is 8 which is 3=blue balls PLUS 5=red balls... and from that you choose 3 balls at random... Because that is what you have to choose at the end of the day according to the question e.g. regardless of you choosing 0=blue balls (3C0*5C3)....@the end you have to choose THREE BALLS. Hence it is 8C3 which is the denominator part (total)
thank you so much :) now i got it ^^
Smanga Sanga
please how did get the numerators and the denominators?
5 choose 3
divided by
8 choose 3
Since you have 5 and you want to choose 3 from them to not get a blue, you say "five choose three" as in "out of five, choose 3 of them". And the same goes for 8 choose 3, except the 8 is the total possibilities you could have chosen (ie. the three blue + five red), but we are still choosing three.
The "X choose Y" stuff refers to the Combination formula.
@@jokerererer I don’t get it, wdym by choose?
5(3) means combinations of 5 taken by a subset of 3 and that equals to
5!/[(5-3)!*3!] = 10
waa really helped mee
كفو ذيب شرح ممتاز
pure talent
God bless you sir!
THANK YOU KIND SIR
Great ..❤❤
I have a very quick question about finding P(x=1).
Why can't you do the multiplication rule to find P(x=1)? I thought doing (3/8) * (5/7) * (4/6) would give you P(x=1). Why wouldn't it work for this case?
+brandong Since we do not care about the order of the selection, we are better off using combinations. Looking at your tentative solution, you are implicitly considering an ordered selection: your calculation is really saying a blue ball first, then a red ball and finally another red ball (let us call this b,r,r). BUT, you can also have r,b,r and r,r,b. You can see from this that in an ordered setup, P(X=1) = P(b,r,r)+P(r,b,r)+P(r,r,b) = (3/8)*(5/7)*(4/6)+(5/8)*(3/7)*(4/6)+(5/8)*(4/7)*(3/6), which yields the same answer.
thanks alottttt
Why do we need to use nCr when we have 3 slots for red/blue balls? Why don't we just have 8 combinations such as: RRR, RRB, RBR, RBB, BBB, BBR, BRB, BRR?
But if order doesn't matter, then we should have: RRR, RRB, RBB, BBB
8C3 would give us combinations that repeat such as RRR.
What if there were 30 blue and 50 red balls? Then your first suggestion becomes completely unmanageable.
To understand why combinations must be used, imagine numbering the balls in this case from 1 to 8: say 1,2,3 are blue and 4,5,6,7,8 are red.
Hope this helps.
how you get the 10/56 can you explain please
how did you get the 10/56?
rose Decatur Hi Rose, it is a classic result of combinatorics that the number of ways of selecting (unordered selection and without replacement) r elements from a set that contains n elements is given by nCr = (n factorial)/( (n-r)factorial * (r)factorial). The expression (n)factorial, usually denoted by (n!), stands for the product of the first n positive integers, so for example (5)factorial = 5! = 1*2*3*4*5=120. I hope this helps!
+slcmath@pc can you explain, why it is use combination? why not permutation? :)
thanks man
good job
As a possible professor, I can communicate with you to solve a misunderstanding question
sir plz upload videos of thermal engineering
Can you explain like I'm a baby how on earth you got 56
I suggest you watch the following two videos on permutations and combinations: ruclips.net/video/vuIKbjoKegg/видео.html and ruclips.net/video/Z0rezieX1G4/видео.html. It should clear everything up.
slcmath@pc ni
it doesn't ,I can see the way you got 56 but how did you get 10?
Tobia Charles subscribe to my videos, I explain well
If you want the ez way theres a C and P shortcut for combination and permutation. I know its super late but eh
tysm
at 11.31 , why P(X=x) times E(X)?
I wish I knew the calculator commands >.< It's making me super lost. I might have known them a few years ago but I've been out of math for 2 years now and so my TI-84 is like a maze
You may want to go with a simpler calculator model, say a basic Sharp. It goes for around $20 and should do the trick just fine.
+slcmath@pc i would but im in college and my class requires it.
Can I ask you sir about random variable problem, thanks in advance.
ok, got lost finding the numerator and denominators
then you're stupid af
when u get p(x=0)
why u choose (5c3)
thanks..
+الفارس الأسمر X=0 means that we are choosing zero blue balls. Since we are choosing 3 balls in total, we are therefore choosing 0 balls from the 3 blue balls and 3 balls from the 5 red balls, hence (3C0)*(5C3).
I don't understand! Why wouldn't you show us how you got 10/56?
he just did it one at a time..
(3C0)(5C3) = 10
(8C3) = 56
joebelle dancel thanks 👍🏾
How to compute it, to get the 10
how did you get 1 out of 56 on the 4th probability?
Jardell Delco check my videos sir I give good explaination
How to get 10/56,30/56,15/56,1/56 ???? Answer me plssss i need to understand for coming exam
You do need a basic understanding of combinations. You can check out my web page for statistics: web2.slc.qc.ca/pcamire/201-510-LW.html and watch the videos under section 5.
your not explained how you get those numbers
how the 10/56 come from
before learning probability, learn permutation combination
How do u get 10/56?
I used the theory of combinations.
Guys how did he get 56
What you need is basic knowledge of permutations and combinations. I suggest you visit my web page: web2.slc.qc.ca/pcamire/201-510-LW.html and watch the videos of section 5. Afterwards, things will make a lot more sense.