Current Events > Math nerds, get in here

(+) Yeah! (+)

50%

Crash posted...

Do your own damn homework.

---

BroodRyu posted...

50%

---

At least 2

---

Mr Hangman posted...

2/3

There are six balls, six possible draws, three of them are eliminated by knowing you picked green. Two of those three options have the 2G bag.

I think this is correct.

Let G_1 be the event that the first ball in the box is green, and G_2 for the second ball. Then we want the conditional probability P(G_2|G_1), which equals P(G_2 and G_1)/P(G_1).

P(G_2 and G_1) is 1/3, because only one of the three boxes allows this. And P(G_1) is 3/6=1/2 because there are three green balls that could have been picked.

(1/3)/(1/2)=2/3

---

Probability is so bizarre.

2
---

BroodRyu posted...

Probability is so bizarre.

It occupies an unfortunate position where it's one of the most unintuitive parts of math, yet it's one of the most (THE most?) important parts of math to our modern society. It's far too easy to fool people with misleading statistics in today's media environment.

---

75%?
---

Captain_Qwark posted...

Red/red box isn't an option

So you picked either green/green or green/red

If you picked gr/gr then it's 100%
If you picked gr/rd it's 0%

So 50% since you don't know which of those two you picked?

No you forget that there are two cases where its Gr/Gr as well as the one Gr/Rd.

So 66%.

its basically the Monty Hall problem presented a different way.

---

Cheese_Crackers posted...

Then we want the conditional probability P(G_2|G_1)

I thought the conditional probability was the probability before either event had occured, as opposed to the probability of G_2 occuring once G_1 has already occured. They're at different points along the time axis, so to speak

I'm not an expert though, so I'm willing to admit that I'm very probably wrong
---

Everything is 50/50 because it either happens or it doesn't.

---

The cat is dead

---

50%. Theres an even amount of green and red balls.

---

2/3.

Third box is discarded, you're gonna pull out one ball out of 3, and there are 2 greens and one red remaining.

---

(1/3)(1) +(1/3)(0.5) = 0.5

---

The easiest way to do this problem is an event tree.

---

This is small enough to enumerate the possibilities.

Let Box 1 = (Green 1, Green 2), Box 2 = (Green 3, Red 1), Box 3 = (Red 2, Red 3)

What are the possible outcomes of "Pick a box at random and then pick a ball at random"?

B1G1
B1G2
B2G3
B2R1
B3R2
B3R3

are all equally likely. Of those, we remove the options where we selected a red ball.

B1G1
B1G2
B2G3

Since these are all equally likely, in the first two cases removing the other ball gives you a green ball and in the third you draw a red ball.

Hence, 66%

---

here we go...

---

teepan95 posted...

I thought the conditional probability was the probability before either event had occured, as opposed to the probability of G_2 occuring once G_1 has already occured. They're at different points along the time axis, so to speak

I'm not an expert though, so I'm willing to admit that I'm very probably wrong

That's more of a physical or philosophical point, because they're mathematically the same. A probability of 0.3, for example, doesn't really mean anything physically. Mathematically we interpret probabilities as being what happens if you could perform infinitely many identical and independent trials, but physically this of course never happens.

---

Cheese_Crackers posted...

and

OR

---

I came up with 66% but I could be wrong. This reminded me of the Monty Hall problem at first.

---

fifteen percent

HagenEx posted...

2/3.

Third box is discarded, you're gonna pull out one ball out of 3, and there are 2 greens and one red remaining.

But you already pulled a ball out. Either the remaining ball is green, or it's red.
The question isn't "what is the probability of pulling another green ball", but rather "what is the probability that the other ball in the same box is green as well".

---

Cheese_Crackers posted...

teepan95 posted...

I thought the conditional probability was the probability before either event had occured, as opposed to the probability of G_2 occuring once G_1 has already occured. They're at different points along the time axis, so to speak

I'm not an expert though, so I'm willing to admit that I'm very probably wrong

That's more of a physical or philosophical point, because they're mathematically the same. A probability of 0.3, for example, doesn't really mean anything physically. The mathematical interpretation is that if you perform infinitely many identical and independent trials, 30% of them will give that result. But physically this is impossible. You might perform 100 trials but only 29 of them give that result. Does this mean the mathematical probability is wrong? No, and that's why the mathematical definition is separate.

But isn't the whole point of conditional probability that the condition changes the probability? E.g. in this case, the probability of G_2 given G_1 is different to G_2 AND G_1?

The former would be 'further' along the time axis, since G_1 has 'already occured'
---

LordFarquad1312 posted...

But you already pulled a ball out. Either the remaining ball is green, or it's red.
The question isn't "what is the probability of pulling another green ball", but rather "what is the probability that the other ball in the same box is green as well".

There are three cases where you can oull out a green ball. Denominator.

There are two cases where there is a green ball that can be removed leaving behind another green ball. Numerator.

2/3 = 66%

And just to satisfy the remaining outcomes;

There is one case where there is a green ball that can be removed but not leave behind a green ball.

1/3 = 33%

There are no other outcomes.

2/3 + 1/3 =100%

---

math isn't real, it's a made up idea by Big Education to get kids to think that they're either dumb or smart.

---

casio_guy posted...

math isn't real, it's a made up idea by Big Education to get kids to think that they're either dumb or smart.

e.g. casio guy

Captain_Qwark posted...

Red/red box isn't an option

So you picked either green/green or green/red

If you picked gr/gr then it's 100%
If you picked gr/rd it's 0%

So 50% since you don't know which of those two you picked?

Yea. And you showed your work. Good job.

---

teepan95 posted...

But isn't the whole point of conditional probability that the condition changes the probability? E.g. in this case, the probability of G_2 given G_1 is different to G_2 AND G_1?

The former would be 'further' along the time axis, since G_1 has 'already occured'

There's no implied causality or ordering in conditional probability.

The difference between P(G_2 | G_1) and P(G_2 and G_1) is that the first doesn't (on its own) represent how likely G_1 is. Imagine if G_1 and G_2 were very rare events but were comorbid. Then P(G_2 | G_1) would be fairly high and P(G_2 and G_1) would be fairly low. That doesn't mean that G_1 caused G_2 or G_1 came first or anything though.

---

Maverick038 posted...

e.g. casio guy

Lol hes not wrong about the first part. Math is an expression, we created the framework ourselves.

The second part? The educational system is definitely formatted to produce worker bees, not entrepreneurs, and inventors, and social indoctrination. So in that sense it is used as a measuring stick for how smart someone is. But his blanket statement is incorrect.

---

N3xtG3nGam3r posted...

But his blanket statement is incorrect.

more of a cozy sweater statement; blankets are too bulky.

---

divot1338 posted...

There are three cases where you can oull out a green ball. Denominator.

There are two cases where there is a green ball that can be removed leaving behind another green ball. Numerator.

2/3 = 66%

And just to satisfy the remaining outcomes;

There is one case where there is a green ball that can be removed but not leave behind a green ball.

1/3 = 33%

There are no other outcomes.

2/3 + 1/3 =100%

Not really. You pulled out a green ball, that is one of the premises of the problem.
The next ball you take will either be red or green. Only 2 cases.

---

Ignore the troll, it's 2/3rds.

---

Did TC choke on the green ball and die? Or was this a hit and run?

---

Sariana21 posted...

Did TC choke on the green ball and die? Or was this a hit and run?

These are always "hit and run" topics. Make the topic. Walk away. Let the idiots troll each other to a 500 topic. The TC isn't actually interested in the answer to the problem. That's genuinely not the point of the topic.

FWIW, the correct answer is 2/3. But problems like these aren't formulated to be "solved" but rather to illustrate just how bad the human mind is at comprehending probability and statistics in general. You're not supposed to find the answer, you're supposed to find out where the flaws are in the logic used so that you can avoid or account for those flaws in more important problems.

---