Short cut to solve this Probability problem | 𝗝𝗘𝗘 𝟮𝟬𝟮𝟰, 𝟮𝟵𝘁𝗵 𝗝𝗮𝗻 𝗦𝗵𝗶𝗳𝘁 𝟭
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- Опубліковано 29 січ 2024
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There are times when you can answer a question without solving, based on concepts, here this problem from JEE Main 2024 29th Jan Shift can be answered simply without actually solving the problem.
Let's learn how to find the right answer for this question both conceptually and by solving.
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Gp + probability
My approach
Let's consider that
we get 2 in Try(2) so the probability of getting 2 p(2) is 5/6*1/6
Try(4) -P(2) 5/6*5/6*5/6*1/6
Try(6) -P(2) 5/6*5/6*5/6*5/6*5/6*1/6...
..Now P(2)=P(T2)+P(T4)+(PT6)+(PT8)......
Now we got an infinite GP with a=5/36 r=25/36 sum of infinite GP =a/1-r
We get ans 5/11
Nice
Good Afternoon sir
The question on Same concept was asked in last year jee adv.
I paste on chat gpt and chat got solved it.
There I found an alternate way. 😅
Seeing the options if I take the no. of trails 11, the event of getting 2 in even toss will be caused by "2,4,6,8,10" numbered tosses. So the probability is 5/11
yes
Bahuuut accha kiye ho. Jindagi main khoob aagey barhoge.
Very elegant way
@@anmold5676 😅
This is also in ncert example questions in probability chapter
Probability of odd > probability of even I have not understood sir . Can you please repeat it clearly in next video . Please sir .
because it is possible to get 2 on dice on the first attempt and 1st attempt is odd (no. 1 is odd )...
aur ye series infinity tak bhi chalti hai to odd pehle occur ho raha hai to uski probability badh gayi...
There I found an alternate way. 😅
Seeing the options if I take the no. of trails 11, the event of getting 2 in even toss will be caused by "2,4,6,8,10" numbered tosses. So the probability is 5/11
Scenario: you have rolled die two times. Probability of two occurring on first try = probability of two occurring on second try = 1/2
But, if 2 was rolled in first try itself, there is NO SECOND TRY because game stops when 2 is rolled. So then, probability of it occurring on first try increases
for infinitely many tries, probability of rolling on any odd number like 1 will always be more
@@tusharrao74 Bro you know that probability= favourable/ total.
As in the question you can see that the no. of trails is given 11. So, why don't you take this opportunity of taking no. of trails as 11.
This basic convention may solve this question is seconds.
@@amazingcraft6818oh wow that's a great perspective
Sir actually i hadn't studied probability and this question came in my shift and i actually solved like this 😅
2:30
Sir each roll of a die is an independent event,so you must not conclude that if 1st roll is odd numbered roll then it's probability is greater than even numbered roll just because its turn is coming before others.
Please correct me if If I am wrong I repeat my objective that is "you can't establish relation between both independent rolls"! Explain plz...
Sir itne easy question ku exam mai aisa idea dhunde se acha hai solve karna
Sir +4 ho gye confirm😀
Find the values of a, b, c and x, y, z with the value of ...
(We let that...)
k=x×b⁴-b³-b²+b-2(z×c+x×a)×b°
Where we have an equality as,
a×x+y+b×y-z+c×z = {b-[x×b⁴-b³-b²+b-2(z×c+x×a)×b°]}÷ (2×y)
as well as we also have that,
b²=a×c , y²=x×z , y²=b²/2
and....
1). a, b, c belongs to natural numbers.
as well as...
2). x, y, z also belongs to natural numbers.
and...
3). k belongs to real numbers.
Sir please answer and solve my question 🙏🙏
I am you're big fan ❤️
i literally did this in exam only lol
Sir black book ke functions chapter ka question 4 pls solve karein
amgm laga
sir divide by 11 nahi ho sakta(class 8) delhi
It is extremely easy question.
Class 12 boards ka bhi padhaiye sir
Don't worry guy's or
Never give up guy's ✊
I'm nda asparit ✊
Thomsop guy's ✊✊
Also me nda aspirant 😊😊😅
@@abhii793 nda ke sath bsc first year chal rahi hai regular 😀
Bhai tumhara english dekh ke lag raha tumhara hona muskil hai.. But i wish ki ho jaye tumhara nda me❤
@@ayushkumarjha7005 hen 😕
@@ayushkumarjha7005 Q yr 🤕
method1 op maza aagya
Sir please. do pnc question as well.. 8 identical books 4 identical shelves
+1
0008
0017
0026
0035
0044
0116
0215
0314
0413
0512
0611
0422
0323
0125
2222
These are the 15 cases possible for that question and here no arrangement is possible as they all are identical. If shelves would be different then we will apply the beggar's method to solve.
@@Yug1001 Can we do this by beggers method or distribution method or we will have to think the numbers of ways like this only?
@@kisalaymauryavanshi2290 No bro beggar's method are applied for those problems in which beggar's are distinct. For ex here shelves are treated as beggar's which are identical so we have to count cases like that only.
@@Yug1001 thanks 👍 is there no other solid method for these types rather than countibg
Sir please tan√x+cot√x
I did it within seconds even by solving by correct method.
Sir aap kaha se hai 😊😊😀🙏
😅
lim x tends to infinity[x(2x - [x^3-x^2+1]^(1/3)-[x^3+x^2+1]^(1/3))]
I need the answer to this question, I have tried a lot but am unable to solve it.
the limit of the given expression as
�
x tends to infinity is infinity.
@@shreyanshsingh1513 nope the ans is 2/9
I’m doing this in university i feel bad for high schoolers having to do garbage like this
Tukka marne ki Ninja technique, btw yeh meri shift mai aaya tha aur maine kuch aise hi logic se answer tick kiya😂
Meri shift me bhi aise hi question aya tha aur pehli baar me hi 3 options eliminate ho gaye the kyuki wo sabhi 1/2 se kam the
Bilkul bhi sochna nahi pada us sawaal me😂
Mera sahi hai😊
Aur kis kis ka sahi hua?😅
This is actually an NCERT level question....Answer should come out in 5 seconds
bhai kuch bhi 5 sec mein kese nikal dega time lgega kam se kam 30 sec
I am spend my 10 min on this q but not solved 😢😢
MIT LEVEL ENGLISH
PROFESSOR VIVEK SINGH
I did this in the same way 😂😂
First view
Shitt galat ho gayaaaaa 😭😭😭😭