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How P(X=2) =1/4?
Ответитьfor x=2 the first assumption that is 2/5 means 2 always uses so how you assumed it as 1/2
ОтветитьWhen 2 people always use Wi-Fi, then p(x=2) should be 1 right? Why is it .25?
ОтветитьExcellent explanation, thank you!
Ответитьgreat explanation I've never thought of that before whether physics has something to do with heads and tails
ОтветитьThankyou sir
Ответитьshouldn't it be pmf instead of pdf?
ОтветитьHello, sir, thank you for the explanation! But I highly suspect that you are an expert in chemistry according to your profile image. lol
ОтветитьThankyou sir
ОтветитьGreat example!
ОтветитьHello sir, firstly thank you for such great videos.
I have a request though, can you make a video where you just put the formulas used to solve numericals so that students like me can relate the "explanation" to what is asked in exams!
Thanks!
I appreciate that you explicitly state that we know that P(heads) = P(tails) = 0.5 BECAUSE OF physics. I've never thought of that before, but it's really helpful to think of it in that way. There's always a reason behind even the most seemingly obvious things. Thank you!
ОтветитьVideos help me a lot, thanks!😀
ОтветитьSir, can you elaborate more why P(X=2)=1/4, P(X=3)=1/2, P(X=4)=1/4? I was lost on the use of assumptions. Thank you.
ОтветитьSir, can you please make a video on Maximum Likelihood and Maximum A Posteriori decoding?
ОтветитьJust one quick question, How can we associate probabilities with uniformly distributed random variables X= {0,1}, in a string of bits. In other words, If I wanted to send a string of bits (that are generated uniformly) over some channel, then how can I get its respective probabilities that sums to one? A little help would be appreciated.
Ответитьso sir are saying that giving numerical analogy to probability evnet is called random variable
ОтветитьSir ...can you please make a video on DTFT , DFT and FFT
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