JEE Mains: Probability | JEE Live Sprint | Unacaddmy JEE | IIT Mathematics | Sameer Sir

Unacademy JEE brings you another JEE Maths session to prepare you for JEE Mains 2020. In this session, Sameer Sir will be discussing JEE Mains important questions of Probability and tips & tricks to solve questions and will help you revise the topic according to JEE Mains 2020. Watch this video to know some of the best shortcuts to solve JEE Mains question paper.
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Learn the complete details about Probability for IIT JEE and related Concepts in this video. It is one of the most important topics of the IIT JEE Maths question paper - in IIT JEE Advanced and JEE Mains as well. Get the best IIT notes curated by experts. Solve important JEE Main and JEE Advanced 2020 questions along with IIT JEE maths preparation tips and tricks. Crack IIT JEE 2020 Maths with these valuable IIT JEE preparation tips and tricks.

Sameer Sir is a part of the alumnus of IIT Bombay. Being an IITian himself, in this IIT JEE Maths Live Online Class, Sameer Sir will touch upon introduction, important questions, and strategy for Probability. He is a passionate teacher and has helped many students crack IIT JEE. During LIVE online classes he shares his personal experiences with students preparing for IIT JEE 2020. This helps the students get acquainted with the IIT JEE exam and make a study plan and strategy that will help them crack IIT JEE mains and advance in 2020.
So, are you ready to crack the JEE Main 2020 Exam?

The dates aren't final because of the coronavirus outbreak, but the IIT JEE 2020 exam will happen soon after. There’s not much time left for preparations for IIT JEE Maths. The time is now when you have to buck up and start preparing and revising all important concepts of questions for JEE Mains and Advanced. If you act now, your dream of making it into IIT can come true.

In this video we aim to prepare you for JEE Mains Maths 2020 (JEE Maths) and answer your following queries:
— JEE Mains Maths Preparation tips
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— How to revise for JEE Mains | Time table to study for JEE Mains
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[ishanmisra]

There are four ways in which he can write books.
Here, G - Good Book & B - Bad Book
GG, BB, GB, BG
P(atleast one book published) = 1- P(none book published)
So, for 4 cases :
P(none book published) = (1/2 * 1/3 * 1/2 * 1/3) + (1/2 * 3/4 * 1/2 * 3/4) + (1/2 * 1/3 * 1/2 * 3/4) + (1/2 * 3/4 * 1/2 * 1/3)
=169/576
P(atleast one book published) = 1 - 169/576
=407/576
Option (A)

[sumitsouravchoudhury]

P( none of books published) =
(1/2*1/3 + 1/2*3/4)(1/2*1/3 +1/2*3/4) =169/576
1-P(none of books published) =
1-169/576= 407/576

[prateekkumar]

hw ques ka ans A option ...p(atleast 1)=1-p(none)
P(none published)= 4 cases
{both good not published (1 way)
one good one bad not published (2 ways)
both bad not published (1 way)}
so [solved by total prob. theorem] p(none)=169/576
so ans=1-p(none)=407/576=option A

[thetravellingengineer]

The best mathematics teacher I have ever seen. Thank you Sir. Love ❤ from Kolkata

[avronilbanerjee]

407 /576 option A
Because first calculating
Probability of publishing each book
(probability of publish & bad) + (probability of publish & good)
= (0.5 * 0.66 ) + (0.5* 0.25)
11/24
Now making cases
case 1 1st book is published 2nd is not
case 2 2nd book is published 1st is not
case 3 both are published
calculating case 1
= Probability of publsh(for 1st) * Probability of not publish (for 2nd)
(11/24) * {1 - (11/24)}
143 / 576
calculating case 2
Same ans 143/576
calculating case 3
= Probability of publsh(for 1st) * Probability of publish (for 2nd)
11/24 * 11/24
= 121 / 576
Adding 3 cases we get
(143+143+121) /576
407/576 OPTION (A)

[mukulgarg]

H/w question answer
A option
Done by binomial distribution

[priyanshchaturvedi]

I have watched lectures of many teachers but they are not as efficient as you are in explaining the questions ❤️❤️❤️❤️👍...

[govindkushwaha]

Pahle not publish ki nikal lo
1/2×1/3+1/2×3/4

Phir 1-p(none publish)
1-13/24×13/24
=407/576

[sachalhablani]

Kal ke homework Question ka ans

1 case
Both books are good

½ *2/3 + 1/2*1/4

11/24*11/24=121/576

2 case

Any of the book is good


11/24*(1-11/24)*2

11/24*(13/24)*2

286/576

286/576+121/576

407/576

Ans a
Type karne me bahut samaye laga

[krishnananddubey]

Answer to homework-
Phle probability of publishing book total probability theorem se aa jaega which is = 1/2*2/3+1/2*1/4= 11/24..
After that applying the concept of binomial prob..
P(at least 1)= 1-P(none)= 1-169/576=407/576..
Ans.A

[wisdomkhan]

The thinking of unacademy is really impressive. The last minute pep talk was awesome sir. And also revolutionary. No doubt you have so many subscribers. Your teaching technique is also very impressive sir. Thanks unacademy for this motive.

[harshtyagi]

Sir,
Answer of HW question is A.
I used the tree flow approach and found all ways of publishing 1 and 2 books.
Or we can also do 1-P(none of the book is published). That would be smaller and faster approach as well.
Please confirm the answer

[zahoorahmad]

Probability of getting one book to be published =1/2 ×2/3 +1/2 ×1/4 =1/3 +1/8 =11/24 (because written book can be either good or bad)
Now, probability (atleast one book book to be published)=Probability (exactly one of them to be published or both of them to be published)=probability (first one is published and 2nd one is not published or first one is not published and 2nd one is published or both are published)=11/24 × 11/24 ( because probability of a book to be published =11/24
Therefore, probability of a book not being published =(1-11/24)


Therefore, required probability =11/24 ×13/24 + 13/24 × 11/24 +11/24 ×11/24 =143/576 +143/576 +121/576 =(143+143+121)/576 =407/576 Ans.

[sashidhar]

HW question binomial probability se bhi hog ga na.

[Shivam-choudhary]

1 options is correct
Fiirst of all we find total probability of publishing book
Then applying binomial distribution formula to get 407/576

[gautamvaja]

A
by subtracting possibility of not publish

[aviralvaish]

the title is wrong
it is written unacaddmy

[lakshmipriya]

37:50 total possible matrices should be 7 to the power 6 why it is 7 to the power 9 as diagonal elements must be zero. So answer should be C. I think sir is wrong.

[nidhishukla]

Spelling of unacademy in name of video is wrong

[kaladeviwarkade]

Sir I'm in class11 and i got a supplementary in chemistry..toh mai kese padhu..kaya 11class ka supplementary paper necert se aata hai kaya..