Why one rope is not suffcient?

Fold it half and then again fold it. You will have four equal parts. Time taken to burn 3 parts is 45 minuts. is it not?

]]>Anyway, the correct answer to question 7 is 1/2, the water level in question 8 does not change (as already stated in an edit at the bottom, but it would be nice if this was more clearly indicated in the answer itself instead of first giving the wrong answer and then correcting it), and plane 2 only gives a quarter tank of fuel to plane 3 (which is a third of what it has, not “one-half”)

]]>P(same sex) = P(two boys AND boy seen) P(two girls AND girl seen) = P(two boys | boy seen) * P(boy seen) P(two girls | girl seen) * P(girl seen). This has to be 1/2, obviously, but would be 1/3 if the answer to question 7 were 1/3 (which would be false).

The mixed drawer of course had a black and a white sock, I don’t know where that red sock came from

And “assuming that plane 2 can refuel at base in literally no time at all” is not required, in fact it has plenty of time to refuel. Plane 1 can leave while plane 2 is refueling, give half its fuel (1/4 tank) to plane 3 at 270 degrees, and then plane 2 can meet them on the way back at 315 degrees with 3/4 of a tank left, just enough for the three of them.

I might also add the following equation:

P(mixed | girl seen) = P(girl seen | mixed) * P(mixed) / P(girl seen) = (1/2) * (1/2) / (1/2) = 1/2

And one last comparison:

Suppose you have an audience of 1000 women, all of which have 2 children. You ask them to pick one of their children at random, and then raise their hand if that particular child was a girl. 500 hands go up, since they are obviously just as likely to pick a boy or a girl. 250 of these have two girls, which is 1/2.

Now ask the audience to raise their hand if they have at least one girl. This time, 750 hands go up, a third of which has two girls.

Question 7 is equivalent to the first situation, not the second.

The answer would be 1/3 if you met the woman at work during “bring your daughter to work day” (with a limit of one daughter per parent).

]]>As others have already pointed out, answer 7 is completely wrong. The funny thing is that you actually give a link to an online discussion which arrives at the same conclusion: the answer is 1/2 (disregarding a few people who just stated their opinion without good arguments). A few “experts” first claimed 1/3 but then changed their minds. You can look at it in two ways:

1. What are the odds of a random family having two children of the same sex? 1/2, right? How does that change if you happen to know the sex of one of them? It doesn’t, obviously, it’s still 1/2. Just like the fact that seeing a black roll at roulette doesn’t change the odds of the next roll. If the chance of a second girl would be 1/3, then the chance of a second boy after seeing a boy would be 1/3 as well, and the chance of a random family having two children of the same sex would also be 1/3 (which is false). P(same sex) = P(two boys | boy seen) P(two girls | girl seen).

2. You are quite right in stating that, after seeing a girl, the children can either be B-G, G-B or G-G. But a mother with B-G or G-B has only a 50% chance of taking her girl out (instead of her boy). So even though there are twice as many mixed families, each has only half as much probability of showing their girl that day. Using Bayes’ theorem, P(two girls | girl seen) = P(girl seen | two girls) * P(two girls) / P(girl seen) = 1 * (1/4) / (1/2) = 1/2

You can also compare with the well-known sock drawer problem: if there are three drawers, one with two white socks, one with two black socks, and one with a black and a red, you draw one sock at randon, it happens to be white, what are the odds the other is white as well? Here the answer is 2/3 while your logic would have given the incorrect 1/2. Your problem is equivalent with four drawers of which two are mixed (correct answer is then 1/2).

Then, question 8, you found out for yourself but it would be nice if you updated the text above. The level does not change.

Finally, question 13, plane 2 only gives a quarter tank to plane 3, not “one-half its fuel”. It then returns with half a tank while plane 3 has a full tank. The rest of the problem is correct (assuming that plane 2 can refuel at base in literally no time at all).

]]>Your brain, thoughts and decisions are just as much part of the universe as the boxes and ultimately driven by the laws of physics, biology and your past experiences. The omniscient predictor (with a brain the size of a planet) knows all of that and can therefore predict everything, including your reaction to his prediction and the experimental setup.

YOU CANNOT ESCAPE

]]>These words have been used to mean each other essentially since the beginning. Specifically, *masterful* ‘imperious’ dates back to 1400, and *masterly* ‘skillful’ to only 1648, so the former is much older. But we find *masterful* ‘skillful’ from 1425, and *masterly* ‘imperious’ from 1544. So there was no period when the two were clearly separated, and if anything the ‘imperious’ sense is the original one for both words.

The use of *masterly* to mean ‘imperious’ is pretty much obsolete: the OED dates its last quotation in this sense to 1766. But *masterful* ‘skillful’ is ongoing. The OED’s headnote does say that it seems to have declined during the 19th century, but there are still three 19th-century quotations and only four 20th-century ones. If anything is doubtful, it’s the gap between the 1425 quotation and the next one, which is Milton’s in 1641.

In short, we cannot save this usage, for there has never been anything to save. This is just another of Fowler’s attempt to draw lines where there are no lines to draw, and to impose his ideas of clarity on a living language.

]]>