Server time (UTC): 2021-01-21, 04:35 # Simple mathematics

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## Recommended Posts  Solutions?

AAHAHH NUMBERS

is it 1/2 or am i stupid? AAHAHH NUMBERS

is it 1/2 or am i stupid?

Idk man, i say its 2/3, i dont want to post my solution too quick, want to see what others say.

AAHAHH NUMBERS

is it 1/2 or am i stupid?

Idk man, i say its 2/3, i dont want to post my solution too quick, want to see what others say.

How can it be 2/3's when there's 2 gold balls left and the next ball you pick is either the silver one from gold&silver box, or a gold one from the gold&gold box? I hate probability calculations by the way. AAHAHH NUMBERS

is it 1/2 or am i stupid?

Idk man, i say its 2/3, i dont want to post my solution too quick, want to see what others say.

How can it be 2/3's when there's 2 gold balls left and the next ball you pick is either the silver one from gold&silver box, or a gold one from the gold&gold box? I hate probability calculations by the way.

Idk man i hate these calculations aswell, i havnt had mathematics in school for 2 years soon But i have asian origin so its ez. (no offense pls) Seven. Seven.

Did you solve it using bayes theorem?

100% since I do this 1/2. It's like that goat and car question except not tricky.

Edit: Or is it?  Seven.

Did you solve it using bayes theorem?

No, absolute convergence theorem.

1:2

There is always a 50% chance that there is another gold ball inside the box.

Just gonna put this in media..

Pretty much if you have one gold ball it can only be one of two boxes, which form 1 has the other gold and the other doesn't, 50/50 chance. SCIENCE BITCH I still stand at 2/3

If it's not 1/2, its 42 no doubt. I am still trying to explain it to Martin. This is how I try to explain it. but 1/2 is too easy

it looks too easy to be easy

Did more calculations to revise for the probability section of my maths exam on Monday (Cheers Martin)

Probability of you choosing box GG is 1/3 at the beginning, while the probability of you getting a G as the first ball is 1/2. Therefore P(GnG) = 1/3 and P(G) = 1/2

The probability of G given the first ball is G = (probability of G and G)/(Probability of G being the first ball)

This is: P(G|G)=P(GnG)/P(G) , so P(G|G)=(1/3)/(1/2) = 2/3.

The probability the second ball is gold is 2/3. Martin was right.

Thanks for making me revise The normal equation is: P(B|A)=P(AnB)/P(A)

Schrödinger's apples. There are apples but there are also no apples.

Really smart people have trouble reading simple questions. The question is that you take a ball from the same box after your first ball. And really smart people are having trouble making exams with trick questions wich is not fair for them of course.

Did more calculations to revise for the probability section of my maths exam on Monday (Cheers Martin)

Probability of you choosing box GG is 1/3 at the beginning, while the probability of you getting a G as the first ball is 1/2. Therefore P(GnG) = 1/3 and P(G) = 1/2

The probability of G given the first ball is G = (probability of G and G)/(Probability of G being the first ball)

This is: P(G|G)=P(GnG)/P(G) , so P(G|G)=(1/3)/(1/2) = 2/3.

The probability the second ball is gold is 2/3. Martin was right.

Thanks for making me revise How could I forget this? Its not too long ago I had it in school. Haha, shame on me.

Edit: But the question is whats the probability of the second ball to be gold, not that both the balls you pick is gold... I still stand at 2/3 I am still trying to explain it to Martin. This is how I try to explain it.

The thing is, if you pick the middle box there is only a 50% chance of it being gold, but with the left box there is a 100% of it being gold.

Therefore there is another image you missed out on the second step, where the left hand ball was picked and the right hand ball is still in the box.

This gives 2 events where the second ball is gold, and 1 event where the second ball is silver. Schrödinger's apples. There are apples but there are also no apples.

Oh god youve never seen me RP when i think theres Apples in the Tradepost have you?

_________________________________________________________________________

Arcus is smart,

So i tink its 1:2

1/2 i stand by that common sense people.

Did more calculations to revise for the probability section of my maths exam on Monday (Cheers Martin)

Probability of you choosing box GG is 1/3 at the beginning, while the probability of you getting a G as the first ball is 1/2. Therefore P(GnG) = 1/3 and P(G) = 1/2

The probability of G given the first ball is G = (probability of G and G)/(Probability of G being the first ball)

This is: P(G|G)=P(GnG)/P(G) , so P(G|G)=(1/3)/(1/2) = 2/3.

The probability the second ball is gold is 2/3. Martin was right.

Thanks for making me revise The normal equation is: P(B|A)=P(AnB)/P(A)

Uh... *hesitates, then speaks softly*... objection?

The question as stated in the OP is as follows: "If the first ball is gold, what is the chance that the second ball FROM THE SAME BOX will be gold too?"

First ball gold means that it can't have been the third box. It is either the first or the second. One of them has a gold ball and the other doesn't, so the chance is 1/2.

How the question is asked and how you read it can be as important as the actual math.

It is late and I'm pretty sure I made a mistake. ## Create an account

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