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## Comment Re:The other problem posed in TFA (Score 1)981

It simply means that there is a 50% chance that Peter has a brother. Peter's sex is not given and it has 50% chance that it may be a girl. :)

It is given. "At least one of whom is a boy" means that the number of boys is guaranteed to be greater than 0. Therefore, one child is a boy, and I pinpointed him as being the given of the problem, and gave him a name to differentiate him from the other unspecified child. If the problem had stated "Exactly one of whom is a boy", then the probability of both children being boys is 0 (0%), because the number of boys is guaranteed to be greater than 0 and less than 2.

Assuming that all families have exactly 2 children with random sex distribution:

1. Is at least one of your children a boy? - Yes, it is. Then the possibility that your other child is a son is 1/3.

Because 75% of families have at least one boy and 25% have two boys.

You gave the right probability here, but for the wrong reason. "Is at least one of your children a boy?" answered in the affirmative means that you can now answer the question as if one boy was a given. The question now is, "Given that I have at least one son, what is the probability that I have 2 sons?"

Per this page, this can be written as P(2 sons | at least 1 son) = P(2 sons and at least 1 son) / P(at least 1 son) = (1/4) / (3/4) = 1/3.

And I just invalidated all of my other comments on this thread... Ouch!

*takes a huge bite of humble pie*

## Comment Re:The other problem posed in TFA (Score 1)981

And now you're saying that it's twice as likely to have a girl as it is to have a boy?

You say:
P(Peter, Boy) + P(Boy, Peter) = 1/3
P(Peter, Girl) + P(Girl, Peter) = 2/3

Let's try to reverse-engineer this problem.

Would you agree that "I rolled a die and it landed on 6. What's the probability that it landed on 6?" yields a probability of 1 (100%)? That's because it's a given of the problem. We don't even have to know if it's unweighted, or 6-sided; it could be 100-sided, and it still wouldn't change the fact that it landed on 6. If you don't agree with this, then you say "the probability is undefined, because I have insufficient information about your die".

Would you agree that, in "I tossed a coin twice, and at least one of these landed heads. What's the probability of both having landed heads?", a given of the problem is that there was 1 Heads, and therefore that the probability we're looking for is 0.5 (50%) for the unspecified coin? If you didn't agree with saying that a given of a problem has P = 1, then the universe of the problem is {(H,H), (H,T), (T,H), (T,T)}, each occurring with equal probability, and the answer is 0.25 (25%). Note that, if you don't accept givens to problems, (T,T) is not impossible, because you ignore "and at least one of these landed heads".

The problem posed in TFA amounts to "My wife gave birth twice, and at least one of the children was a boy. That boy was born on a Tuesday. What's the probability of both being boys?" Therefore the problem is exactly like the two coin problem, with 1 boy being a given, and the birth weekday being extra information that isn't used in the problem's question therefore doesn't affect the resulting probability. What's the probability now?

## Comment The other problem posed in TFA (Score 1, Insightful)981

Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?

This is the question posed without the birth weekday specified. TFA actually tries to say that there are 4 outcomes for the pair of children, one of which is impossible, so they remove it. Since "boy, boy" is only one of the 3 outcomes, then the probability must be 1/3. Right?

Wrong.

The boy (let's call him Peter) being a boy is a given of the problem, so it has P = 1. The other child -- we don't care about it being born before or after Peter -- is independent, so the probability that it's a boy is 0.5*. The 4 outcomes are as follows:

Peter, Boy = 0.25*
Peter, Girl = 0.25*
Boy, Peter = 0.25*
Girl, Peter = 0.25*

So, whichever way we slice this problem, the solution is 0.5*.

P(Peter, Boy) + P(Boy, Peter) = 0.5*
1 * P(Other is Boy) = 0.5*

- - - - - -
* May slightly differ due to the male:female ratio at birth. It is assumed here to be 1:1.

## Comment Re:It's easy (Score 1)981

I don't know about the chance of it landing heads up, but if you threw it up, it has a rather high chance of being damaged by your stomach acid...

## Comment Re:Ordering and Convergence (Score 4, Informative)981

The problem is stated thus:

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

One of whom is not "exactly one of whom", so 'one' might be opposed to 'the other' or 'two'. All we know is that one of the problem poser's children is a boy born on a Tuesday. It states nothing about the relationship between the two children in time or space, so the probabilities are independent.

Further, the problem doesn't ask about any probability related to the second boy's birthday. The problem doesn't ask, e.g., What is the probability that my other child is a boy not born on Tuesday?. That makes the birth weekday completely irrelevant.

## Comment Re:Ordering and Convergence (Score 1)981

What if the second child was born on Tuesday but was not a twin of the Tuesday boy?

Say, the Tuesday 75 weeks later. It's still a Tuesday...

## Comment Snow? Feh! (Score 1)560

I bet you would freak out if there was snow in California. Hell, the UK was like a bunch of pussies last winter because of about 15 cm of snow (half a foot), using up all of their grit, and Canada gets more than that every winter.

It just depends on what you're used to.

## Comment Re:All growing and no shrinking (Score 1)247

It will probably be increases of 10.8%, 8.5% and 6.8% compared to the previous proportion, not the current entire Internet.

So if violence grew by 10.8%, but the previous proportion was 4.4%, you jump to about 4.9%.

## Comment Re:All true, but (Score 4, Informative)50

The COINS dataset, which is talked about in the summary, was recently published in CSV files (not proprietary), compressed with Zip (not proprietary) and distributed using HTTP (not proprietary) and BitTorrent (not proprietary) at the user's option.

## Comment Re:Look at Firefox as well (Score 1)234

I think that the clearing of private data in Firefox is a bit counter-productive, because deleting from SQLite databases merely marks the rows' storage space as being reclaimable within the file.

I once cleared private data for a day when my places.sqlite was around 70 MiB, then checked the file size and saw that it hadn't even changed by one byte. It wouldn't surprise me if the URLs were still in there -- all of them, intact, until you visit other pages to make Firefox overwrite the reclaimable pages in places.sqlite.

Even if Firefox truncated places.sqlite when the user clicked "delete everything", the URLs would still be readable on the underlying storage device. Firefox would have to shred(1) or zero out the file. I doubt that's going to happen.

## Comment Baselines (Score 3, Insightful)1138

Education and money are very much alike in one aspect: if everyone has at least the same amount, then that amount becomes the baseline, below which it is worthless.

College degrees being required for plumbing jobs and the like are only the symptom of this problem.

Whereas before education was made mandatory in most countries of the world, the baseline was no education at all, now the United States have college as a baseline. And it's rather difficult to get out of this, because you ask someone in college why they're in college and they'll say, "I must, because I can't afford to not keep up with my peers." So people go to college because people go to college, and it's a recursive clusterfuck.

## Comment Re:But it is +1 better (Score 1)532

Or even 36D. But personally I'd rather have 36DD. Mmm...

## Comment or pull them back (Score 1)242

and reuse or recycle the parts.

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