Animal wrote: ↑Sat Mar 19, 2022 11:11 pm
But, the question should be worded:
logrithmic
A^b = 343
B^c = 10
A^c = 7
What is B^b?
No, and here's why: this is a question from a national mathematics olympiad. These questions are designed around two strict ideas, one that you come to the answer as quickly as possible, and two that you use the problem and given facts about the problem as efficiently and fully as possible in accordance with using best math solving strategies. You can still have the right answer and be marked wrong due to how badly the judges see you use the given info or where your strategy forks. These kinds of questions do not have extraneous information, and the information is meant to be used completely towards answering the problem. You are not supposed to re-think or question the information, you don't have time for that shit.
Going back to the original question, you were given the question of what BB is based on the three given equalities. You can solve for the answer without bothering to try to answer what the exact values of A, B, and C are. You don't need to worry about their values, because you can use the given equalities to do that. Can you tell me what number B is that B multiplied by itself B times would make 1000? Equivalently, can you tell me what number B is that [B x log B = 3]? No, but using the given equalities gets you around that.
You said that AB = 343 = 73 means A = 7 and B = 3. That could be true, but it could also be the logarithmic equivalent of dividing by a hidden zero. The valid step is to make the bases both be A on either side of the equation. That way you don't need to know what the value of A ever is directly. But since they are the same base A, their exponents are equal. That's valid, and we never need to concern ourselves with whether A really is 7 or not.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
50-50
nope.
It's not the Let's-Make-A-Deal problem.
Pulling a white ball tells you you don't have the double black bag. Therefore you either have the all white or the white - black bag. Meaning the one ball left is either the white from the double white or the black from the white - black bag thus 50-50.
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
50-50
nope.
It's not the Let's-Make-A-Deal problem.
Pulling a white ball tells you you don't have the double black bag. Therefore you either have the all white or the white - black bag. Meaning the one ball left is either the white from the double white or the black from the white - black bag thus 50-50.
The odds can't be 50/50 if there are 3 balls left. (2 white, 1 black)
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
50-50
nope.
It's not the Let's-Make-A-Deal problem.
Pulling a white ball tells you you don't have the double black bag. Therefore you either have the all white or the white - black bag. Meaning the one ball left is either the white from the double white or the black from the white - black bag thus 50-50.
yeah, but that's not correct. there is only a 1 out of 3 chance that the remaining ball is black.
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
50-50
nope.
It's not the Let's-Make-A-Deal problem.
Pulling a white ball tells you you don't have the double black bag. Therefore you either have the all white or the white - black bag. Meaning the one ball left is either the white from the double white or the black from the white - black bag thus 50-50.
The odds can't be 50/50 if there are 3 balls left. (2 white, 1 black)
If all the balls left were in one bag you'd be right. But they are in two bags and you are limited to the bag you've already picked.
Animal wrote: ↑Thu Mar 24, 2022 9:31 pm
There are three bags on a table.
In one bag there are 2 white balls.
In another bag there are 2 black balls.
in the last bag there are 1 black ball and 1 white ball.
Now without knowing which bag is which, you grab any bag and pull out a ball. That ball is white. What are the odds that the remaining ball in your bag is white?
50-50
nope.
It's not the Let's-Make-A-Deal problem.
Pulling a white ball tells you you don't have the double black bag. Therefore you either have the all white or the white - black bag. Meaning the one ball left is either the white from the double white or the black from the white - black bag thus 50-50.
The odds can't be 50/50 if there are 3 balls left. (2 white, 1 black)
If all the balls left were in one bag you'd be right. But they are in two bags and you are limited to the bag you've already picked.
yes, you are right. you are limited to the same bag you drew the first ball out of. You can't draw the second ball from a different bag. 2 out of 3 chance your remaining ball is white.
Bag 1 with 2 white balls: after pulling one white ball, the chance of the second white ball is 100% (1 in 1).
Bag 2 with 2 black balls: after pulling one black ball, the change of a second white ball is 0% (0 in 1).
Bag 3 with 1 white ball and 1 black ball: this is somewhat trickier.
If you pull the white ball, the chance of pulling another white ball is 0%, (0 in 1)
but if you pull the black ball, the chance of pulling the white ball is 100% (1 in 1)
So the average chance with this bag is 50% (1 in 2).
The chance of picking any bag is 1 in 3, or 33%, so that has to be factored in.
So the chance of getting a second white ball after picking one bag at random and pulling a first white ball is:
(1/3) x 100% + (1/3) x 0% + (1/3) x 50% = 33% + 0% + 17% (slightly rounded up) = 50% overall chance of pulling a second white ball.
I will fully admit statistics was a really weak area of my math studies, though.
If you can't be a good example, you can still serve as a horrible warning.
“All mushrooms are edible. Some even more than once!”
これを グーグル 翻訳に登録してくれておめでとう、バカ。
QillerDaemon wrote: ↑Fri Mar 25, 2022 2:08 pm
The way I'm seeing it:
Bag 1 with 2 white balls: after pulling one white ball, the chance of the second white ball is 100% (1 in 1).
Bag 2 with 2 black balls: after pulling one black ball, the change of a second white ball is 0% (0 in 1).
Bag 3 with 1 white ball and one black ball: this is somewhat trickier.
If you pull the white ball, the chance of pulling another white ball is 0%, (0 in 1)
but if you pull the black ball, the chance of pulling the white ball is 100% (1 in 1)
So the average chance with this bag is 50% (1 in 2).
The chance of picking any bag is 1 in 3, or 33%, so that has to be factored in.
So the chance of getting a second white ball after picking one bag at random and pulling a first white ball is:
(1/3) x 100% + (1/3) x 0 + (1/3) x 50% = 33% + 0% + 17% (slightly rounded up) = 50% overall chance of pulling a second white ball.
I will fully admit statistics was a really weak area of my math studies, though.
You have to take into account what knowledge you have going into the problem and what knowledge you gain along the way.
Going in, you know that one bag has two white, one bag has two black, and the other bag has both.
But, then after you draw out the first ball and see that its white, you know have that knowledge, which only tells you that you have eliminated the bag with 2 black balls as the bag you chose. So, that's where the question starts. With all of this knowledge, what are your odds that the remaining ball in your bag is white?
I will try to explain why 50:50 chance is wrong and 2:3 chance is right.
All that you know at this point is that the next ball you pull is either white or black and that you have eliminated the bag with only black balls in it.
Now, imagine if each of the white balls was marked with a tiny number that no one realized was on them. One ball marked 1. One ball marked 2. And one ball marked 3. We will call them 1W, 2W, and 3W. The 4th ball is B.
Now, in one bag were balls 1W and 2W. In the other bag were balls 3W and B. All you know is that you drew out a white ball, but you have no idea which one it was.
So, the outcomes possible for the next draw are:
Outcome #1: If you first drew 1W, then you will now draw 2W. Outcome is a white ball.
Outcome #2: If you first drew 2W, then you will now draw 1W. Outcome is a white ball.
Outcome #3: If you first drew 3W, then you will now draw B. Outcome is a black ball.
So you have a 2 out of 3 chance to draw a white ball next.
Animal wrote: ↑Fri Mar 25, 2022 5:15 pm
I will try to explain why 50:50 chance is wrong and 2:3 chance is right.
All that you know at this point is that the next ball you pull is either white or black and that you have eliminated the bag with only black balls in it.
You're right, I get it now. Basically "it's not soup yet!" I was looking at the odds -before- picking the bag and the first white ball. We need to be looking at the odds *after* the picking the bag and pulling the first white ball, when the whole event circumstances have changed, and changing the event's odds. Interesting!
If you can't be a good example, you can still serve as a horrible warning.
“All mushrooms are edible. Some even more than once!”
これを グーグル 翻訳に登録してくれておめでとう、バカ。
If you can't be a good example, you can still serve as a horrible warning.
“All mushrooms are edible. Some even more than once!”
これを グーグル 翻訳に登録してくれておめでとう、バカ。
