Amazing! 💫 Sir,I request you to explain us more Jee Advance Problems I'm a jee Aspirant.(Vector,3D,Probability,Matrices,Calculus etc...) Keep up good work. ❤️❤️❤️
You could also take a shortcut. At 12:30 we know that the number of people who have either 2 or 3 of the symptoms is 30 + 40 + 50 + 60. Subtract that from 900 to get the number of people with either 0 or 1 symptoms, which is 720. Then calculate the probability. It's a good idea to do it both ways to double-check the answer.
A comprehensive lesson both in general terms, and those specific to this problem. Absolutely worth working through - if for no other reason than to see how many mistakes and how many kinds of mistakes you can make if you don't already know how this works. And how to do it correctly. An education in fifteen minutes!
Hi, I was doing this on my own and I can’t understand what I did wrong. the notation i used was f= only fever b=only breathing etc fb= fever and breathing problems fc= fever and cough problems etc fbc= fever breathing and cough problems n= no problems 160= f + fc + fb (alr subtracted the 30 ppl w all problems) 190= c + fc + bc 190= b + fb + bc 330= f + c + fc 340= f + b + fb 670= 2f + c + b + fc + fb 670= 160 + f + b + c one symptom = 510 and then as i try to solve for other stuff i get diff answers and idk why
Group F was with fever, but not only with fever. In F there are also people with other symptoms. Someone may be in two or three groups, but still is just 1 person.
hello first of all, you are right it looks long but the singel steps should be easy i think you only have to bring all the infos in a equention. 3:48 that have to be the first step ok 6:01 i see the problem AND the solution at this point! we check out how many elements are the same and how many not and add those values so becauce ther is no other posible case we have this equantion and it is always true make sense! 7:38 of cause becauce than you have a nice overview! 9:28 ok there it will be hard to understand for me becauce i have never learnd quantity theory! 10:39 ok an normal equantion that should be also posible for me! 12:05 ok thats realy not as difficult as i thought! 13:06 ok the rest is easy 14:40 allright anybody make mistaks! 15:39 ok basicly we devide the sum of this numbers by the count of all people right? 16:42 you found a mistake afterwords :D but that is ok! so now i am a bit more claver! xD 16:49 and your true sentence at the end nice! LG K.Furry
how? can u tell me how u solved it orally i haven't studied 11th probability yet. So if it requires some concept of 11th probability then tell me I will read prob. first and then try it once again
Amazing! 💫
Sir,I request you to explain us more Jee Advance Problems I'm a jee Aspirant.(Vector,3D,Probability,Matrices,Calculus etc...)
Keep up good work. ❤️❤️❤️
Do in Coaching Institute
Sir, could you please make a video covering graph theory? I’m trying to learn it for the amc 10/12.
You could also take a shortcut. At 12:30 we know that the number of people who have either 2 or 3 of the symptoms is 30 + 40 + 50 + 60. Subtract that from 900 to get the number of people with either 0 or 1 symptoms, which is 720. Then calculate the probability. It's a good idea to do it both ways to double-check the answer.
A comprehensive lesson both in general terms, and those specific to this problem. Absolutely worth working through - if for no other reason than to see how many mistakes and how many kinds of mistakes you can make if you don't already know how this works. And how to do it correctly. An education in fifteen minutes!
wonderful to see van diagram and revising it.
8:48 why is this "n(F U C) = 330" and not "n(F U C) + n(F ∩ C) = 330"?
A classy problem ✅
You did more work than needed to solve the problem. Once you had the intersections, just subtract that from 900.
Oh. I was trying to fill all the spaces in case someone needed to learn something.
Always best to show the whole process rather than let the viewer decide fir themselves what happens at the rest of the stages
Sir please make videos on real analysis....pleaseeeeeeeeeeee
Can't we just plug nos into the Venn diagram?
Elements present / Elements common
Hi, I was doing this on my own and I can’t understand what I did wrong.
the notation i used was
f= only fever
b=only breathing etc
fb= fever and breathing problems
fc= fever and cough problems etc
fbc= fever breathing and cough problems
n= no problems
160= f + fc + fb (alr subtracted the 30 ppl w all problems)
190= c + fc + bc
190= b + fb + bc
330= f + c + fc
340= f + b + fb
670= 2f + c + b + fc + fb
670= 160 + f + b + c
one symptom = 510
and then as i try to solve for other stuff i get diff answers and idk why
I disagree. People who have at most one symptom are the 420 people in the Vann diagram. All minus the 480 who have no symptom.
That would be the number of people with at LEAST one symptom, not at MOST one.
@@woodchuk1 You are right! Sorry
This is the simplest form of probability that is taught in 9th grade
: 🤓
Hum sab 11th me Sikh rhe hai aur tuzhe alag se 9th me padhaya
Yeah kid you are the reincarnation of Newton or Einstein
@@KRO_VLOGS tune nahi padha lekin 9th/10th me hota h probability
@@shashankmaurya1263 I think you meant the n(a union b) formula
It looks like inclusion-exclusion principle
if 630 persons are infected, so only 270 people are non infected. am I wrong?
Only 420 are infected do focus on sets union
Group F was with fever, but not only with fever. In F there are also people with other symptoms. Someone may be in two or three groups, but still is just 1 person.
hello first of all,
you are right it looks long but the singel steps should be easy i think you only have to bring all the infos in a equention.
3:48 that have to be the first step ok
6:01 i see the problem AND the solution at this point! we check out how many elements are the same and how many not and add those values so becauce ther is no other posible case we have this equantion and it is always true make sense!
7:38 of cause becauce than you have a nice overview!
9:28 ok there it will be hard to understand for me becauce i have never learnd quantity theory!
10:39 ok an normal equantion that should be also posible for me!
12:05 ok thats realy not as difficult as i thought!
13:06 ok the rest is easy
14:40 allright anybody make mistaks!
15:39 ok basicly we devide the sum of this numbers by the count of all people right?
16:42 you found a mistake afterwords :D but that is ok! so now i am a bit more claver! xD
16:49 and your true sentence at the end nice!
LG
K.Furry
Solution could've been much simpler.
Also I was able to solve it orally😊
how? can u tell me how u solved it orally i haven't studied 11th probability yet. So if it requires some concept of 11th probability then tell me I will read prob. first and then try it once again
@@AadiSrivastava-sp9zn no actually it’s just set theory with simple concept of probability: Favourable/Total
@@ARounDeRoP right
@@ARounDeRoPYeah...
That was fun! Thanks