I doubt any of you mathfags will be able to solve pic related

I doubt any of you mathfags will be able to solve pic related

do yr own homework brainlet

There are n/2 white socks aND n/2 + 1 black socks.

Where 200 < n + 1< 250

A black sock was lost, making the probability equal. Next.

embarrassingly wrong, try harder

Let B be the number of black socks left and W be the number of white socks.

[eqn] P(\text{White}, \text{White}) = \frac{W}{W+B} \frac{W-1}{W+B-1} [/eqn]
[eqn] P(\text{White}, \text{Black}) = \frac{W}{W+B} \frac{B}{W+B-1} [/eqn]
[eqn] P(\text{Black}, \text{White}) = \frac{B}{W+B} \frac{W}{W+B-1} [/eqn]
[eqn] P(\text{Black}, \text{Black}) = \frac{B}{W+B} \frac{B-1}{W+B-1} [/eqn]


[eqn] P(\text{White}, \text{White}) + P(\text{Black}, \text{Black}) = P(\text{White}, \text{Black}) + P(\text{Black}, \text{White}) [/eqn]
[eqn]W (W-1) + B (B-1) = 2 W B [/eqn]

Now you quickly see that 105 black socks and 90 white socks works.

Yeah...This is one of those things that are misleading because of the probability of picking matching sock 50% of the time which doesn't happen if there aren't the same number of socks of each color because the first one you pick might be the missing color, so you have to do some kind of Bayesian thing, but I am too tired so the solution is that since he is a mathematician, no one cares if his socks match anyway, and if he did then he'd keep them is separate drawers, or chose the same color from what is left over, and how often does this guy do laundry if he has 20 fucking pairs of socks anyway, and why are math problems always so unbelievable as if you'd ever need to figure this out in real life, it's Christmas dammit, didn't he get any socks that weren't black or white, or is this guy Jewish, oh shit, now I sound like a racist, I wonder if anyone will read this....

Sorry it's 105 white socks and 120 black socks. [math] W + B [/math] must be a square number since

[eqn] (W-B)^2 = W+B [/eqn]

and 15^2 = 225 is the only square number in the range.

100 white
125 black
he lost a black sock.
May have done calculation mistake but it's merelly geometry at this point. Love this kind of problem.

If X = # of black socks in the two socks picked at random, then X ~ HGeom(2, b, w + b) and

P(X = 0) = 1/2
P(X = 2) = 1/2

That's as far as I get.

Jew man here.

It's not exactly misleading if you actually decide to work the problem. You don't even need to know anything about statistics to find your pattern.

For instance, if you have 2 white socks and 2 black socks, you can just enumerate the socks as W1, W2, B1, and B2. Listing unique combinations:

>You can pick W1 with W2, B1, or B2.
>That leaves W2 with B1 or B2.
>B1 can go with B2.

That's 2 matches out of 6 unique combinations. A 1/3 chance even though 1/2 of the socks match. It's counterintuitive unless you do your simplest-case examples, and you'll find your patterns just by working other simple examples. The first statisticians did it, and you can, too!

oh didn't see your answer and indeed made me realise I don't know my squares, i counted 25 rather than 15 :
you're right, 105 120 thus he lost a white sock

however, I dare say you're quite troubling yourself, you just need to count when you take a not matching pair that is 2WB and say the probability is still 1/2 :

2WB/card( WuB2 \ diag) = 1/2

oh and everything is discrete so prob in this case is just counting measure.

On another note, it's been so long, how do you use latex in here already?

This dude is right

How can this be right.
If there are more black than white, then the chance of randomly drawing isn't 50/50

If I'm thinking about this correctly the probability of getting a getting two black socks does not have to be the same as getting two white socks. It can still average out to 50% chance of getting matching socks when picking them randomly.

Could you explain how you found 105 , 90 as a solution, I lost you on that step

Where did this expression come from?

You rearrange
[eqn] W (W-1) + B (B-1) = 2 W B [/eqn]

2b(b-1)+2w(w-1) = (b+w)(b+w-1)
(b-w)^2 = b+w
15^2=225
w = 105
b = 120
White sock was lost

I'll re write fully so that people see it more clearly :
In the following universe
[math]S = B \cup W \ \ \Delta = \{ (x; x) \in S^2 \}[\math]
[math] \Omega = S^2 \setminus \Delta [\math]
(because we can't take two times the same socks)
the random variable is then : I take a pair of socks (an elt in [math]\Omega [\math]) and determine weither they match or not (0 or 1) :
[math]X : \Omega \rightarrow \{ 0; 1 \} [\math]
of course it's indeed a random variable, because the tribe is the powerset, but let's not write every details. The probability measure is the counting measure (card of set divided by card of total).
Rather than calculating directly, let's calculate the probability of having two mismatching socks :
[math] P(mismatching \ socks) = P(X^{-1}(0)) = 1 - 0.5 = 0.5 [\math]
And you have two mismatching socks whenever the first is black the second is white or the contrary i e : [math] |B| \cdot |W|+ |B| \cdot |W| [\math]
Finally : [math] |\Omega| = (|B| + |W|)^2 - (|B| + |W|) [\math]
thus :
[math] \frac{ 2BW}{(B+ W)^2 - (B + W)} = 0.5 [\math]
[math] 4BW = (B+W)^2 - (B + W)[\math]
[math] (B+W) = B^2 + W^2 - 2BW [\math]
[math] B + W = (B - W)^2 [\math]
everything else has already been said : the sum is between 200 and 250 and a perfect square, and it is the difference in socks.

Sorry ...

probability makes me wanna throw up

105 white
120 black
white was lost