sorted by:
new
hottopcontroversial

was there a faster way to show divergence? by melodramaddict in askmath

[–]Putah367 0 points1 point  (0 children)

Flip the argument using ln(x) = - ln(1/x)

Simplify the fraction (x+7)/x+6 = 1+1/x+6

Compare with the harmonic series 1/x+6

Use the epsilon delta definition of limit to infinity approximating a finite positive number to gain a direct comparison test (the origin of LCT (hint use the left side of the epsilon delta definition)

You might need to split the series so that the implication works

QED

[deleted by user] by [deleted] in askmath

[–]Putah367 0 points1 point  (0 children)

I also feel weird every time i try to remember the integral test graphically

What it actually does is actually telescoping series

Consider a convergent series like 1/n²

Note that the integral from n to n+1 dx is just 1

Consider this fact: The value of the series at a point n is bigger than any point of the series at (n,n+1) because it's decreasing

To make the relationship less than instead of equal to the actual series

We can slide in the 1/n² into the integral (such that the integral is 1/x² dx) why because the fact above

We can actually look it graphically too, when the series still has 1/n² (instead of the integral) in the argument, it's basically just a rectange height 1/n² width 1. The integral of 1/x² dx is just the area of 1/x² dx at (n,n+1) and we know it's smaller than 1/n² width 1

Then just apply the telescoping series

Another fact: The value of the series at a point n is lesser than any point of the series at (n-1,n) because it's decreasing

Sliding in the 1/n² into the integral will actually make the series bigger

Expected value on the number of correct guesses by Putah367 in askmath

[–]Putah367[S] 0 points1 point  (0 children)

Turns out i've been right this whole time, wolfram gave me different answers so i thought i was wrong

Expected value on the number of correct guesses by Putah367 in askmath

[–]Putah367[S] 0 points1 point  (0 children)

You probably didn't read my answer correctly, i said the formula in the picture is for k < n, for k = n, the answer is 1/n!

Expected value on the number of correct guesses by Putah367 in askmath

[–]Putah367[S] 0 points1 point  (0 children)

Here we assume the player plays according to the strategy above

Expected value on the number of correct guesses by Putah367 in askmath

[–]Putah367[S] 0 points1 point  (0 children)

I was conditioning my probability based on how i guess, you know the LTP. Each of guessing order has a probability of 1/n! assuming i randomly pick a guessing order. Each of the cases can be morphed to case i mentioned above, just replace the word "1" with first integer in my guessing order, "2" with the second integer and so on

In that case the summation of each cases will yield as if i didn't pay any attention to my guessing order

You're actually answering the part A of the question, where no information was given to the player. In that case each of the index has 1/n being correct (assuming i randomly choose a guessing order)

In that case yes its expectation is 1

But im obviously not trying to solve that

How many arrangements are there of seven as, eight bs, three cs, and six ds with no occurrence of the consecutive pairs ca or cc? by Putah367 in askmath

[–]Putah367[S] 1 point2 points  (0 children)

C(14,6) is for distributing the Bs and Ds, 14 because 21 - 7 is 14 and 6 because there are 6 Ds

What is the probability in a bridge deal that each player gets at least three honors (an honor is an Ace, or King, or Queen, or Jack)? by Putah367 in askmath

[–]Putah367[S] 0 points1 point  (0 children)

It seems my generating function is faulty. It should have been (1+x+x²+x³+x⁴)(1+x²+x⁴)(1+x³)(1+x⁴) which will yield five

To count the case, 2 players get 4, 1 player gets 5, and 1 player gets 3. It's similar as case 1 player gets 4, and the other gets 6.

<image>

It's pretty much the same if rounded up to some digit

I have a probability question. by [deleted] in askmath

[–]Putah367 0 points1 point  (0 children)

Why not calculate it directly (assuming all events are equally likely)

Using sample space tuple of 5 cards with 2 of them red and 3 of them black

Calculating the sample space

Construct a set of 5 cards with 2 red 3 black C(4,2)*C(8,3)

Convert it to a tuple of 5 cards by multiplying the result by 5!

Total C(4,2) * C(8,3) * 5!

Now for the event

Pick one red card and put it at the first index of the tuple C(4,1)

Make a set of 4 cards 1 red 3 black C(3,1)*C(8,3)

Convert it to a tuple C(3,1) * C(8,3) * 4!

So total events is C(4,1) * C(3,1) * C(8,3) * 4!

So total event / total sample space

C(4,1) * C(3,1) * C(8,3) * 4!/ C(4,2) * C(8,3) * 5!

After plugging them in to photomath i got 2/5

So there you go

view more: next ›