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Attached: A53E9707-1649-4D3E-A791-D25C15396E37.jpg (320x269, 30K)
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50/50
50%
Wrong
1/3, if you already got a gold ball, and you have to reach into the same box for another, there's only 1 of the 3 boxes that could have 2 gold balls.
Wrong
Wrong
How's it wrong? You've already got a gold ball. That means you only had two of the possible boxes. The ball will either be gold from box 1 or silver from box 2.
One out of three, if each box only contains two balls then the only thing that matters is the box you pick.
33%
20%
2/3
2/3
This
Correct
Correct
/thread
Wrong.
Right.
Technically its dependent on how you account probability. The true answer is either 100% or 0% but it is an unkown factor as we cannot see all variables (it being the box which variable has technically been removed from the overall probability now).
However if we include the boxes it would be 50%. Though as stated, our outlook on probability changed because we had a new set of variables which changed the odds from the original set.
In mechanical engineering I like to call this "an argument over semantics."
There are 3 ways to get a gold ball in the first pick, all equally likely. First way is ball one from box one, second way is ball 2 from box 1, third way is 2nd box. If you ended up with box one, (66% chance) you get another gold ball when you take the second
As I said earlier.... wrong
monty hall does not apply. 50% for this
If you think something is wrong but you can't explain why then you're probably the one who's wrong.
If we number the balls in the picture 1, 2 and 3, we ended up with 3 Scenarios:
>You picked Gold Ball #1; Other ball (#2) is gold
>You picked Gold Ball #2; Other ball (#1) is gold
>You picked Gold Ball #3; Other ball is silver.
So there's a 2/3 chance the other ball is gold.
In this scenario, the first ball is already drawn, meaning the scenario is really “If you have two boxes with a gold ball and a silver ball, what is the probability you draw a gold ball”
50/50
Correct that monty hall does not apply here but incorrect answer
fucking engineers
I bet you're not even smug about this post. It's just "The Right Answer" cause you're a failed mathematician and a trained mentat.
Wrong
Wrong. That fact you picked a gold ball first has an impact on your second pick....
1/2, if the first ball is gold then we know it isnt the 3rd box, of the boxes that could provide the first gold ball one has a gold ball remaining, the other does not
>The ball will either be gold from box 1 or silver from box 2.
so its a 50% chance the next ball after you picked up the 1st gold ball will be gold/silver. thats why its not 1/3 (33.3333%)
You do not have a second pick. Read the OP as many times you have to until you understand how you were wrong.
Wrong
well i do not understand. pls explain
There are 6 possible choices for the first ball, all equally likely (probability 1/6)
Box1Ball1
Box1Ball2
Box2Ball1
Box2Ball2
Box3Ball1
Box3Ball2
But we picked a gold ball, so there are 3 possibilities, also all equally likely (probability 1/3):
Box1Ball1
Box1Ball2
Box2Ball1
Now, if we picked Box1Ball1 or Box1Ball2, the next ball is guaranteed to be gold (probability 1)
If we picked Box2Ball1, the next ball is guaranteed to not be gold (probability 0)
So the total probability of getting a second gold ball, given that we saw a first gold ball is:
(1/3)*1+(1/3)*1+(1/3)*0 = 2/3
Or you can use Bayes theorem, but we're on Sup Forums so we have to assume nobody here has a math education beyond middle school.
Correct
explain. if something is one of two options it has to be a 50-50 chance, given that the boxes were selected at random
Wrong
2/3.
This is a trick question, which is why so many people are struggling with it.
This is a case of the OP not comprehending the way their own question is phrased: It would be 2/3 only if you could pick a different box. But because only 2 boxes have gold balls, and you have to pick from the same box twice after already getting a gold ball, then that means that there's a 50% chance the box would contain a second gold ball. Box 3 is irrelevant as it contains 0 gold balls, and per the phrasing of the question the box we picked already had 1 gold ball, and we can't switch boxes so therefore the silver box doesn't factor into the odds. The first gold ball came from either box 1 or 2, therefore it's a 50% chance. OP is autistic.
1 in 3. 33%
troll....the probability in this question is 1/2. the third box is discounted at the point of calculation.
50%
The gold ball came from either left or center.
There are two possibilities;
* next ball is silver (first gold was from the center box)
* next ball is gold (first gold was from the left box)
If we picked a gold ball (which we did) it is twice as likely that we are in box 1 than box 2. As we know there are no gold balls in box 3 so can eliminate it. This means there’s a 66% chance we’re in box 1 and a 33% chance were in box 2. If we’re in box 1 that next ball has to be gold and if we’re in box 2 the next ball has to be silver. Given the 66% chance were in box 1 the chance of the next ball being gold is 66% or 2/3
Oops, found a moron who doesnt understand probability lmao. Theres a correct answer.
It is always used in calculation, the 3rd box was nevee removed. Remember it was 1 in 3 before you reached into the boxes, your odds do not change now that you learn more about the variables because they always remain consistant
7/9
Maybe this will help:
Let's say you have 2 boxes. One box has 1 billion gold balls, and 1 silver ball. The other box has 1 billion silver balls, and 1 gold ball. You pick a ball from one of the boxes at random, and it ends up being gold. Which box do you think it's more likely you picked from?
Seven of nine, or seven ninths. 14/9/2
This
This, anyone who doesnt understand needs to go back to school. Also fuck the engineer who can't even use his brain in this thread earlier.
50% and I refuse to believe otherwise because I am based
It is one half chance. Any other formulation needs better wording of the problem.
Wrong
The silver box doesn't matters, since you picked a gold ball, hence you have either picked the first or second box. You'd think you have a 50% chance of having picked either box, but think again, information has been revealed. There are 2 gold balls in one box and only 1 in the other, the chances of having picked the first box are thus 2/3, since you already know you have a gold ball.
Also I'm gay because I touched balls.
Since we are dealing with a hypothetical question and asking for probabilities, the real question is are we dealing in quantum mechanics or not?
Because your answer will carry based off of this.
Wrong
>calling self based
>using that as an excuse for ineptitude
Lurk more newfag. You need to learn how this works.
a box with a gold ball or more has been selected.
there are two of these boxes.
in one box, the other ball is a silver ball. in the other box, the other ball is a gold ball.
there are two outcomes. it is a 1/2 chance to receive a gold/silver ball.
if you were to pick a box at random and then pick the 2nd ball out of the box, it would then become a 1/3 chance to receive a silver ball.
what the question is asking for is the first answer, 50%.
This is one of those like the Monty Hall Problem.
No, we aren't. These are solid gold balls.
The correct answer is 2/3
If you dont understand, you shouldn't comment on quantum mechanics.
Yes. In that it's basic probability, and trivial to arrive at the correct answer, but the majority of people are morons and get it wrong.
Wrong
No. It's not.
Its 2/3. 66.6666%
can be in 3 boxes 1 1 1, or 2 1 0. the troll in the red text is we are not supposed to know what we can see
Red text says we cant see inside the boxes, but we can see we have a gold ball that we pulled out. That means that it cant possibly be box 3. 50%
Wrong
wrong
Sorry correct not wrong
No, it means there’s a 50/50 chance that you’re either in box one or box two. All you know is that you’ve drawn a gold ball, which is available from box one or box two.
Now you have a probability of picking up either A.) a gold ball, or B.) a silver ball. Those are the only two options that you could pick up. your probability is 50/50.
The question isn’t “what’s the probability of getting both gold balls now that you’ve got one gold ball”. The question is “what is the probability that the NEXT BALL is gold.”
we have to assume a random distribution for the troll to make sense
0,3
2/3 propabilty to get a box with a golden ball
1/2 propabilty to get another golden ball after you eliminated one box by finding it with a p=2/3
its "conditional probability" P(B|A)
B=Finding another golden ball
A=Finding one golden ball among three boxes
so its 2/3 * 1/2 = 0.3333...
Believe me....You are wrong
You're right, it can't possibly be box 3. But why do you think boxes 1 and 2 are equally likely? Box 1 has more gold balls, so if you picked a box at random, and happened to pull out a gold ball first, it's more likely that you picked box 1 than box 2.
everboy who wrote 0.5 or 50% never went to school or learned the concept of conditional probability
cond. probabilty basic stuff its 1/3 or 0.3333
There are actual 2 p conditions which are connected by a causal step ... condition 1 after condition 2...
Absolute BASIC math stuff.
except what remains in the two possible boxes is only one ball in each, and that one ball possibly in the boxes is either gold or silver, not gold, gold, or silver because there's only two balls left you could possibly pick out of the one and only box you are pulling a ball from
You already have a gold ball. The probability of getting a box with a gold ball is 1
right.
Yes, but you are not counting the probability from the start, you are counting the probability from step 2 - the point of knowing 1 of the balls
Wrong
but the chance that you got that golden ball in first try has its own probabilty p=2/3..
This doesnt make sense.
2 is more than one, there are objectively more gold balls in one of the 2 boxes you could've taken from
Thus you probably took from that one, and not the other one.
right! basics
every steps has its own p
You didn't read the OP, user
No, that's how it was meant to be interpreted.
You already know one ball is gold at step one.
I just did, what are you going to do keyboard warrior?
the questions is about the final probabilty after taking one golden with p= 2/3 so its conditioned probabilty they will never explain this in tests because its too basic stuff
If we didn't pull out a ball already probability to pick a 2xgold box is 33%, but since we know 1 ball is gold its 50%
Wrong
Except you skipped a step.
You already know one of the balls is gold, your chance of taking a gold ball from any of the 3 individuals was .33, once you took the ball out, and found a gold ball, your likelyhood of getting another one doubles, as you are more likely to have pulled from the box with 2 balls, and eliminated a move.
You now have a 2/3 chance of pulling another one, as you'd pulled the first one and wasted away your 1/3
conditional probability is partly right
you forgot that there are 2 golden balls..
so one can be empty after first try
using all info from pic...correct in 50%. ignoring what it tells you to ignore is not.
Ooh feisty.
>I'm wrong and also willfully misinforming people about something I clearly dont understand
2/3 = first probabilty
1/3 * 2/3 = second probabilty
(meaning: if the golden originated from the mid box then there are one filled and two empty boxes left p = 1/3 chance to get another
if the golden ball originated from the left box then there are two filled boxes left and p is
2/3 chance to get another on second try
)
2/3 * (1/3 * 2/3) = 0.66
first stage 3 gold, 3 silver - 1/2 - second assuming random allocation 2/5 - we do not know where the balls are
fucking this
Hmmm...So he was behind that...?