/sci/ - Science & Math

>Can you do it?
Your homework? Bayes' Theorem. Look it up.

50%. He either did or he didn't.

zero dumbass

0%.
A didn't receive answer.

(n-1) / (2n-1)

50% his reply may have been lost in the mail

>>2174337
but what if b got the letter but his REPLY was lost in the mail?

(2n-1)/(n^2)

>>2174345
THe answers obviously has to be in terms of N, you fuckin' idiot.

>>2174357
That's wrong.

>>2174358

OK, 2n then if you want to be all scientific about it

2n is the correct answer
n it was lost in transit A to b
n the reply was lost in transit
n+n or 2n

(n-1)/n

P(B):1-1/n
P(~A):1/n+(1-1/n)*1/n = 1/n+1/n-1/n^2 = 2/n-1/n^2
P(~A|B):1/n
P(B|~A) = P(B)*P(~A|B)/P(A)
=(1-1/n)(1/n)/(2/n-1/n^2)
=(1/n-1/n^2)/(2/n-1/n^2)
=(n-1)/(2n-1)

the proability that B gets his letter is
1-1/n since that is the chance its lost in the mail.

>trollface.jpg

1/n are lost in the mail
1 - 1/n are not lost in the mail

answer: 1 - 1/n = (n-1)/n

>>2174374
retard

Either he got it or he didn't. It's 50%.

1/n chance that B did not receive that letter. Period.

However, if we're considering the possibility that B did receive the letter and his reply was lost in the mail.....

((n-1)/n)(1/n)=(n-1)/n^2

Wow. FOr such an easy Baye's Problem a lot of people have no clue.

>>2174384

Oops, I forgot the return probability.

(n-1)/n that B got it

(n-1)/n x 1/n that his reply was lost = (n-1)/n^2

1/n that the original was lost.

1/n + (n-1)/n^2 = (2n-1)/n^2

Not very hard.

0%

We know for a fact that he did not receive the letter

>>2174392
brofist.jpg
>>2174381 here

n/(2n - 1)

>>2174395
Saying it's not hard while failing to get the answer makes you look like an ass. What you've calculated is simply the probability of A not receiving a reply. (P(~A) in this solution >>2174381)

>>2174392
>implying the problem stated more than it did

>>2174399
Keep on keepin' on, brotha'!

i think a few people assumed that B would write back if he didn't receive a letter

>>2174412
How so?

>>2174418
How can you write BACK to something you haven't received?

>>2174421
exactly

yet a few of these solutions seem to based on something like that

>>2174421

>Dear A,
>
>Where the fuck is my letter?
>
>Yours,
>B

>>2174420
>one in n letters are lost in the mail
>what is the probability that B recieves the letter

if it stated that you waited for a reply and got none, you would have to account for the fact that it was lost on the way back, but not as stated here.
Also not the same as 1 - A recieves reply

OP here.

Interesting problem, huh?

Later tonight I will post another interesting one to which I have an answer that disagrees with an answer I found on the internet from my textbook's publisher; but I know mine is the TRUE and HONEST answer. Stay tuned for that question, coming up after the break.

>>2174433

What's so hard?
If B got the letter, he would have replied.
So if A gets no reply it can mean one of two things:

> the letter got lost on the way to B
> or B's reply got lost on the way to A

>>2174443
>reading comprehension fail

>>2174421
People answering 1-1/n are assuming that the letter didn't get lost going from B->A, since 1-1/n, is the overall probability that B received the letter. What you need is the probability that B received the letter GIVEN that A did not receive a reply, which is why you need to use Bayes' Theorem.

WOOW EVERYONE HERE IS RETARDED...

IT'S: 2/n

Really think about it. For fuck sakes.

N/(n-n)

>>2174450
You'll have to elaborate.

>>2174409

Just because I rushed/didn't check my answer doesn't mean the question isn't easy, thus I don't look like an ass.

Given A didn't receive a reply with probability (2n-1)/n^2, we can find the probability received the letter by finding the part of this probability which has his return letter being lost, because he always sends a letter back if he receives one.

The probability of his return letter being lost was (n-1)/n^2

((n-1/)n^2)/(2n-1)/n^2 =(n-1)/(2n-1)

Not very hard.

>>2174433
Not him, but he did it right. The question asks for the probability that B received the letter GIVEN that A did not receive an answer.

OK< HERE!

1/n+(1-1/n)1/n

>>2174459

>one in n letters are lost in the mail
>what is the probability that B recieves the letter

>one in n letters are lost in the mail
>what is the probability that B recieves the letter

>one in n letters are lost in the mail
>what is the probability that B recieves the letter

If A sends a letter to B, there is a 1/n chance it is lost

the chance it is not lost therefore is
1-1/n

>>2174465
>hard enough to take you two tries

whether B wrote back or not is irrelevant

1 in n letters get lost in the post

the chance A's letter arrived is n-1/n

>>2174474
see
>>2174453

>>2174453

Never heard of Baye's theorem and I arrived at the correct answer. (n-1)/(2n-1). My solution is the one in 3 parts with a lot of text. Just use common sense.

>>2174316
Let S be the event B receives a letter and T be the event A does not receive a letter.
Then we want:
<div class="math"> P(S\mid T)=\frac{P(S\cap T)}{P(T)} .</div>
The probability of S and T is given by the probability of B receiving a letter, <span class="math"> \frac{n-1}{n} [/spoiler], multiplied by the probability of A receiving a letter given that B has received a letter, namely <span class="math"> \frac{1}{n}.
Thus <div class="math"> P(S\cap T)=\left ( \frac{n-1}{n} \right )\left ( \frac{1}{n} \right )=\frac{n-1}{n^2} </div>
The probability of T can be found from the probability of the complement event, namely the probability of BOTH A and B receiving letters. This is simply <span class="math"> \left ( \frac{n-1}{n} \right )^2 [/spoiler].
Thus <div class="math"> P(T)=1-\left ( \frac{n-1}{n} \right )^2=\frac{2n-1}{n^2}. </div>
Finally we have:
<div class="math"> P(S\mid T)=\frac{P(S\cap T)}{P(T)}=\frac{\frac{n-1}{n^2} }{\frac{2n-1}{n^2}}=\frac{n-1}{2n-1} </div>
Note: This answer makes sense, since if n=1, we would have the case where 100% of all letters are lost in the mail thus the desired probability should be 0.[/spoiler]

Suppose n^2 letters get sent to B. n^2/n = n letters are lost, which means that n^2 - n replies are sent to A. Of those n^2 - n replies, (n^2 - n)/n = n-1 letters are lost, which means that A receives n^2 - 2n + 1 letters.

For the remaining 2n - 1 letters that A sends but does not receive a reply to, where did they get lost? N letters got lost from A to B, and n-1 got lost from B to A. B received the letter and A didn't only when the letter got lost from B to A. A never received a letter 2n-1 times, and of those 2n-1 times B received a letter n-1 times, so the odds are (n-1)/(2n-1).

Now its time for me to look at the thread and laugh at those who fucked it up.

>>2174475

One half-hearted, 30 second attempt that I knew was wrong, so not really a 'try' at all.

Obvs if I had this question in an exam I'd actually think about it for more than 2 seconds before diving into an answer that I knew was overlooking something.

>>2174316
Let S be the event B receives a letter and T be the event A does not receive a letter.
Then we want:
<div class="math"> P(S\mid T)=\frac{P(S\cap T)}{P(T)} .</div>
The probability of S and T is given by the probability of B receiving a letter, <span class="math"> \frac{n-1}{n} [/spoiler], multiplied by the probability of A receiving a letter given that B has received a letter, namely <span class="math"> \frac{1}{n}.[/spoiler]
Thus <div class="math"> P(S\cap T)=\left ( \frac{n-1}{n} \right )\left ( \frac{1}{n} \right )=\frac{n-1}{n^2} </div>
The probability of T can be found from the probability of the complement event, namely the probability of BOTH A and B receiving letters. This is simply <span class="math"> \left ( \frac{n-1}{n} \right )^2 [/spoiler].
Thus <div class="math"> P(T)=1-\left ( \frac{n-1}{n} \right )^2=\frac{2n-1}{n^2}. </div>
Finally we have:
<div class="math"> P(S\mid T)=\frac{P(S\cap T)}{P(T)}=\frac{\frac{n-1}{n^2} }{\frac{2n-1}{n^2}}=\frac{n-1}{2n-1} </div>
Note: This answer makes sense, since if n=1, we would have the case where 100% of all letters are lost in the mail thus the desired probability should be 0.

>>2174467
>wheredoesitsaythat.jpg

either way, the probability is equal both ways, so you need to consider Not A, and A, Not B if you want to consider the probability it is lost,

>>2174327
+1

>>2174453
Bayes' Theorem helps formalize it, but if you know the right populations to divide with each other (for probability of A given B, divide all the times where A and B occur together by the number of times that B occurs)

I sincerely hope I am not embarrassing myself by fucking up this explanation.

>>2174496
>>wheredoesitsaythat.jpg
>inthefirstsentence.pcx

>>2174520
aaaaannnnnndddd I look like a retard, since I forgot that Bayes' theorem has multiple formulations, of which one of them I stated in plain English. And I only remembered off-hand the formulation that used P(B|A) to calculate P(A|B)

>>2174524
>gardenpathparagraph.png

>>2174520
That's basically just a restatement of Bayes' Theorem.

>>2174524
>stilljustsasksforoneway.jpg

>same probablilty each way so not hard to figure out the >probability it is lost on the way back which is what you >are doing.

>>2174537
>pokerface.jpg

The lack of a reply is irrelevant to the question.

(n-1)/n

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>>2174472

Man, look at these trolls.

Nice one OP. Don't think people going on about conditional probability and return letter were trolls. They just fell for your clever question.