0:00 Sharing a birthday in class room 10:05 Monomers in a peptide chain 30:00 How many times should I toss a fair-sided die such that probability of getting 2's is greater than two-thirds? 30:42 Probability of getting at least two 5's in three tosses in a fair-sided die 35:36 Probability of at least two 5's
35:29 I think the answer would be 16/6^3. This is because question said "at least" so, there is another case where all are 5. Please correct me if I am wrong.
last problem in another way: there are 3 possible configs of two 5's:- 55x, 5x5, x55 now x can have 5 values 1234&6 which means each configs when taking each of the values individually will produce 5 sample points. Hence there will be 5*3= 15 sample points hence p(2) = 15/6^3
No, Poor is the one who does not know the joy of problem solving. And this is college dude he can at least expect from his students to know basic combinatorics.
0:00 Sharing a birthday in class room
10:05 Monomers in a peptide chain
30:00 How many times should I toss a fair-sided die such that probability of getting 2's is greater than two-thirds?
30:42 Probability of getting at least two 5's in three tosses in a fair-sided die
35:36 Probability of at least two 5's
Thank you
Thank-you so much for this lecture series. Because of the very clear explanations, I was able to get a distinction for advanced stat mech
The book followed in the lecture series is Meheran Kadar statistical mechanics of particles.
It's important* the sequence of this lecture is wrong,please follow from NPTEL site to get the proper sequence.
@29:10 (ln(1/3))/(ln(5/6)) is easier to solve than hit and trial, no?
i think he didn't want to use log as many might not be familiar with it
8:30 you forgot the factor of sqrt(2πn), although in this case it doesn't matter much because you're dividing by 300!.
Thanks sir
35:29 I think the answer would be 16/6^3. This is because question said "at least" so, there is another case where all are 5. Please correct me if I am wrong.
Yes the answer is 16/6^3 which simplifies to 2/27. Prof missed that case of 555.
last problem in another way:
there are 3 possible configs of two 5's:- 55x, 5x5, x55
now x can have 5 values 1234&6
which means each configs when taking each of the values individually will produce
5 sample points. Hence there will be 5*3= 15 sample points
hence p(2) = 15/6^3
Thnx sir
i am getting ans of burthday prob as 99.5.... where am i doing mistake
the name of the book who are you refer
please where is lecture 2 , thanks
here's the whole playlist ruclips.net/video/8xRFqrNyJCg/видео.html
when you speak, leave out the words "okay" and "basically"
Technically, he does not understand the meaning of math and do math using formula! Poor his students
No, Poor is the one who does not know the joy of problem solving. And this is college dude he can at least expect from his students to know basic combinatorics.
@@rishabhkhatri yes, you are right.