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  • 14 votes cast
Apr
6
revised Prove that $\frac{f(n)+a}{g(n)+b} = O(\frac{f(n)}{g(n)})$
added 127 characters in body
Apr
6
asked Prove that $\frac{f(n)+a}{g(n)+b} = O(\frac{f(n)}{g(n)})$
Jan
10
comment Notation of functions
Could you please explain more about this? Why would anyone need to do this? Perhaps you could provide a link to some reference? Also, could you explain the usage of "|" symbol in definition? What does it signify? Or is it interchangeable with ","?
Jan
10
asked Notation of functions
Nov
4
awarded  Yearling
Jul
2
awarded  Curious
Jun
13
comment Sum of normally distributed random variables
Thanks, I didn't think about this
Jun
13
asked Sum of normally distributed random variables
Jun
12
comment Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?
Binomial distribution: E[X] = np, if E[X] = 3 and p = 0.3 we get n=10 and the number of failures is 10-3=7. Isn't it strange that this gives the same answer even if this is not in binomial distribution (as far as I understand)?
Jun
12
comment Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?
But I am aware of Pascal's distribution which looks similar to NB. Using that I get that E[x]=n*0.3/0.7 = 3 => n=7. So the number of failures is 7-3=4?
Jun
12
comment Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?
@Studentmath no, I am not
Jun
12
asked Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?
Jun
11
asked Probability of event happening in Bernoulli scheme if most likely outcome is given?
Jun
10
asked What events should be added for it to become sigma algebra?
Jun
10
accepted 5 cards are drawn. Probability of second or third card being diamonds.
Jun
10
comment 5 cards are drawn. Probability of second or third card being diamonds.
So, if I understand correctly, then the answer simply is 1 - (39*38)/(52*51) ?
Jun
10
asked 5 cards are drawn. Probability of second or third card being diamonds.
May
13
asked Random variable and CDF
Mar
7
awarded  Notable Question
Mar
6
accepted Continuous probability. Angles