₦airaland Forum

Obamaofusa:
That is the measurement we are using now until Google(who is neutral)gives another.Onitsha is a mere 52 square km.

That's what I'm saying.
Google only measures the urban areas.

Example is this Google Earth map of ibadan.

Iyanaoffa, akanran, Olodo.

They are part of ibadan, yet google didn't measure it.

Obamaofusa:
Ibadan is 3080 square km according to Google.

Guy, reason, how can Ibadan be 3,080 when Oyo state is 28,000+.

Google only tell you the urban areas.

Kinematics:
The Nsukka thing is a lie naa. It's not something for debate.

Okay.

Kinematics:
No verified official data claims any city in Nigeria is more than 3500 km sq. None! Let's stop the lies on nairaland.

No official data, but we can verify it ourselves.

I schooled in UNN Enugu.

It doesn't take me 2yrs to know everywhere in Enugu (capital )

No, official data but it was claimed the largest city and 2nd in Africa by World statistics

Kinematics:
This thread is 95% lies 5% truth. The op started the lies so I dont blame the others.

Even tiny states like Anambra, Imo, Abia and Enugu dey claim over 3000 km sq cities. Others in the west claiming over 9000 km sq for Ibadan. After these lies all of una go still wan go heaven.


Chisos Chisos!!

Actually Ibadan is over 9000km²
It's larger than Some South East states.

If you add the while South east states together it's a bit bigger than Oyo state.
What then makes you think Ibadan only won't be bigger than some of their states.

Like Enugu, Enboyi, Anambra
Each of the states above are less than 9,000km²

If you stay Ibadan for 20yrs, you can't know everywhere. It's not possible.

Are you telling me I can't know the whole Enugu or The whole Anambra or the whole Enboyi in 5yrs.?

That is actually less than Ibadan by far?

Anambra is not up to 5 LG in Ibadan.
Enboyi, Enugu are not up to 9 local government each in Ibadan.


This is Ibadan (capital) Vs some south east state (state + capital)

Edit your post and copy the content then put link.

Else it won't be accepted

ElectroLyte:
https://thestreetjournal.org/2020/05/meet-bolaji-agboola-gambari-president-buharis-chief-of-staff-son/

THE STREET JOURNAL
www.thestreetjournal.org

ahaz:
but your oluwole n.b.s statistics stated that your erosion ravaged brown roof hamlets is the richest in Nigeria how come your people are stating other wise or is there something you are not telling us?

Maybe you need glasses to read the report.

When there was lock down, who went to farm

Where you able to go to farm?
Oh, sorry you guys don't have farm self.
Erosion has destroyed your lands.

You don't know what it means.

APCLyingBastard:
I told you that you will enter depression.

It is not scientifically proven

APCLyingBastard:
Did anybody ask you the probability of picking a gold ball first?

You see why I said comprehension and a narrowed thinking set was your problem.

Okay, I give up.

APCLyingBastard:
Stop thinking linearly.

The only thing you are to consider is which of the boxes with at least one yellow ball did you first draw from.

The question is supposed to be from your second draw.

But the question says I can't see the box?

What's is the probability that I picked the Gold ball self, at first.

APCLyingBastard:
sammy07, has it hit you hard now!

lol

Here is another way to view it.

It's not 1/2 because you are more likely to have picked the left box than the middle one in pjhase 1, given that you got a gold one to begin with. Had you picked the middle one in ph1, half of the time you wouldnt be able to proceed with the problem because you'd have grabbed a grey one.

There are 6 outcomes, all equally likely, in part 1 - you pick:
Left box, left ball
Left box, right ball
Middle box, left ball
Middle box, right ball
Right box, left ball
Right box, right ball

Since you know you achieved gold though, we know only 3 of those outcomes could actually have happened:
Left box, left ball
Left box, right ball
Middle box, left ball

If you picked either of the left box options, you will grab the other in the second phase, which is gold. If you picked the midde box in ph1, you will grab a grey ball in ph2 and fail. 2 of those 3 possibilities result in success - so the odds are 2/3

APCLyingBastard:
Ok sammy,

Your problem is you are counting balls when you are not to.

Instead of 1 + 2+ 3= 6

See it as oranges and apples or let's say A + B.

You really need to think outside the box - literally and not concentrate on what is in the box.

You have eliminated one of 3 boxes by drawing a gold ball.

You are now left with 2 boxes.

The chance that you drew from either the box with two gold balls or the one with a silver and a gold ball is what the next probability outcome is all about!

Jeez!

Sir, we're talking about probablity of picking gold ball.
You shouldn't label box instead you should label the balls.
Well have two gold balls in a box, while will you then label the box.


There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

So, I still disagree that it's 1/2

APCLyingBastard:
[s][/s]

When you see that the probability is not in the picking the ball but the box you chose, that is when the answer will hit you.

By picking the gold ball, you eliminated the box containing 2 silver balls.

You are now left with two boxes.

You either drew from the first with two gold or the one with a silver and a gold ball.

Forget about the number of balls and think about the box.

Here lies the logic.

Forget about the box and think of the balls.
We are talking about balls here not box.


As others have noted, it's a deliberately confusing question. They label the boxes. What you need to do is label the balls.

There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

When in doubt, list out all the possible outcomes and count 'em up.

APCLyingBastard:
[s][/s]

According to your pics, you marked only the second box.

Why ?

Why wasn't it picked in the 1st box.

Olodo.
See how I labelled it here with detailed explanations.
OLODO

As others have noted, it's a deliberately confusing question. They label the boxes. What you need to do is label the balls.

There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

When in doubt, list out all the possible outcomes and count 'em up.

APCLyingBastard:




Let me put you out of your pathetic ignorance and misery.

let me provide the LOGIC behind the answer which is 1/2 or 50%.

You have 6 balls , 3 gold and 3 silver arranged in 3 boxes with 2 gold, 2 silver and one of either in separate boxes.

You draw a gold ball.

The question that was asked is what is the probability of you drawing a second ball from the same box?

That means you drew from either the box with 2 gold balls or the one with a single gold and silver ball.

Now you have only 3 balls left in two boxes where you had gold balls

The chances of you having drawn a gold ball from the box with either 2 gold balls or a single gold ball is inconsequential as the main question is the follow-up event of drawing a second ball.


Hence, you are left with deciding if your first draw was from either the box with 2 gold balls or the one with a single gold and a single silver ball.

The probability that you will pick a gold ball is 50% because if you drew from the first box with 2 gold balls , it would have been 100% but then again you might have drawn from the second box with a single gold ball and a silver ball.

So you are either going to pick a Gold ball (if you selected from the first box) or a silver ball (if your first draw was from the second).

Olodo.
OLODO
Go to school you no gree

As others have noted, it's a deliberately confusing question. They label the boxes. What you need to do is label the balls.

There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

When in doubt, list out all the possible outcomes and count 'em up.

BabaRamota1980:
Its 2/5

The guy nah Olodo

OLODO

As others have noted, it's a deliberately confusing question. They label the boxes. What you need to do is label the balls.

There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

When in doubt, list out all the possible outcomes and count 'em up.

falcon01:
1/1 and 1/2 = 2/3

The guy nah Olodo

OLODO

As others have noted, it's a deliberately confusing question. They label the boxes. What you need to do is label the balls.

There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

When in doubt, list out all the possible outcomes and count 'em up.

APCLyingBastard:
[s][/s]

Keep it coming.

I see how you zombie brains are wired to failure.

Olodo, You don't have anything answer to refute it.


If nah your papa born you, prove this wrong.


OLODO rabata, Oju eja Lo MA je, oni m'owema

As others have noted, it's a deliberately confusing question. They label the boxes. What you need to do is label the balls.

There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

When in doubt, list out all the possible outcomes and count 'em up.

[s]

APCLyingBastard:
[s][/s]

When I provide the answer and reason behind it, you will enter depression. [/s]

OLODO

As others have noted, it's a deliberately confusing question. They label the boxes. What you need to do is label the balls.

There are 6 balls; Box 1 contains balls A and B, both gold. Box 2 contains C and D; C is gold, D is silver. Box 3 contains E and F, both silver.

You picked a ball. It's gold. It could be Ball A, B, or C. You don't know which.

If it's Ball A, the other ball in the box is gold.
If it's Ball B, the other ball in the box is gold.
If it's Ball C, the other ball in the box is silver.

2 out of 3 possibilities, the other ball is gold.

When in doubt, list out all the possible outcomes and count 'em up.

APCLyingBastard:
[s][/s]

Olodo that thinks it's 1/2

Refer to this https://en.m.wikipedia.org/wiki/Bertrand%27s_box_paradox
Olodo

APCLyingBastard:
[s][/s]

Comprehension is your problem

It's 2/3rds. You picked a gold ball, which is can be one of three balls with equal probability. The left gold ball in the left box, the right gold ball in the left box, or the gold ball in the middle box.

In two of those three cases the other ball in the box is also gold.

Refer to the Monty Hall problem which has a very similar premise.

APCLyingBastard:
[s][/s]

Comprehension is your problem

It can be easier to understand the correct answer if you consider the paradox as Bertrand originally described it. After a box has been chosen, but before a box is opened to let you observe a coin, the probability is
2
/
3
that the box has two of the same kind of coin. If the probability of "observing a gold coin" in combination with "the box has two of the same kind of coin" is
1
/
2
, then the probability of "observing a silver coin" in combination with "the box has two of the same kind of coin" must also be
1
/
2
. And if the probability that the box has two like coins changes to
1
/
2
no matter what kind of coin is shown, the probability would have to be
1
/
2
even if you hadn't observed a coin this way. Since we know his probability is
2
/
3
, not
1
/
2
, we have an apparent paradox. It can be resolved only by recognizing how the combination of "observing a gold coin" with each possible box can only affect the probability that the box was GS or SS, but not .


https://en.m.wikipedia.org/wiki/Bertrand%27s_box_paradox

You're not wiser than the person who owns it. Lol

APCLyingBastard:
[s][/s]

Comprehension is your problem

The problem can be reframed by describing the boxes as each having one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (), one a silver coin on each side (SS), and the other a gold coin on one side and a silver coin on the other (GS). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold?

The following reasoning appears to give a probability of
1
/
2
:

Originally, all three boxes were equally likely to be chosen.
The chosen box cannot be box SS.
So it must be box or GS.
The two remaining possibilities are equally likely. So the probability that the box is , and the other coin is also gold, is
1
/
2
.
The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically:

The probability that would produce a gold coin is 1.
The probability that SS would produce a gold coin is 0.
The probability that GS would produce a gold coin is
1
/
2
.
Initially , SS and GS are equally likely {\displaystyle \left(\mathrm {i.e.,P()=P(SS)=P(GS)} ={\frac {1}{3}}\right)} {\displaystyle \left(\mathrm {i.e.,P()=P(SS)=P(GS)} ={\frac {1}{3}}\right)}. Therefore, by Bayes rule the conditional probability that the chosen box is , given we have observed a gold coin, is:

{\displaystyle \mathrm {P(\mid see\ gold)={\frac {P(see\ gold\mid )\times {\frac {1}{3}}}{P(see\ gold\mid )\times {\frac {1}{3}}+P(see\ gold\mid SS)\times {\frac {1}{3}}+P(see\ gold\mid GS)\times {\frac {1}{3}}}}} ={\frac {1}{1+0+{\frac {1}{2}}}}\times {\frac {\frac {1}{3}}{\frac {1}{3}}}={\frac {2}{3}}} {\displaystyle \mathrm {P(\mid see\ gold)={\frac {P(see\ gold\mid )\times {\frac {1}{3}}}{P(see\ gold\mid )\times {\frac {1}{3}}+P(see\ gold\mid SS)\times {\frac {1}{3}}+P(see\ gold\mid GS)\times {\frac {1}{3}}}}} ={\frac {1}{1+0+{\frac {1}{2}}}}\times {\frac {\frac {1}{3}}{\frac {1}{3}}}={\frac {2}{3}}}
The correct answer of
2
/
3
can also be obtained as follows:

Originally, all six coins were equally likely to be chosen.
The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS.
So it must come from the G drawer of box GS, or either drawer of box .
The three remaining possibilities are equally likely, so the probability that the drawer is from box is
2
/
3
.
Alternatively, one can simply note that the chosen box has two coins of the same type
2
/
3
of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type
2
/
3
of the time. In other words, the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?".

Bertrand's point in constructing this example was to show that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result; and the two methods are equivalent only if this probability is either 1 or 0 in every case. This condition is correctly applied in the second solution method, but not in the first.

https://en.m.wikipedia.org/wiki/Bertrand%27s_box_paradox

APCLyingBastard:
[s][/s]

I think the answer is 2/3, since the fact that you picked a Gold ball implies a 2x likelihood that you’re in the box than the GS box.

The other way to look at this is to realize that the first pick has 6 distinct balls that you can pick. The fact that the balls are in 3 boxes doesn’t affect the 1/6 likelihood that you will pick any one of the 6 balls. Of course, the second pick is dependent on the first pick.

Let’s say the balls are organized as such: [1, 2] [3, 4] [5, 6], or [G G] [G S] [S S]. You have picked one of balls 1, 2, 3. So, the possible second picks are 2, 1, 4 two of which are G and one is S.


Tell us the answer, but don't ever say it's 50%

I think the answer is 2/3, since the fact that you picked a Gold ball implies a 2x likelihood that you’re in the box than the GS box

With replacement or without replacement?