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It also depends on precisely how well you do in each exam.
Let's say that the C boundary is 60-70%, and the D boundary is 50-60%.
Person A could get 67% (C) in one test and 55% (D) in another, and their average would be 61% (C).
Person B could get 62% (C) in one test and 54% (D) in another, and their average would be 58% (D).
Nuh uh! I got that point fair and square.
Ah, sneaky. I got all twisted up.
Ah, nice work phrontister.
I didn't participate in this one too much, but I do enjoy these, gurth. Keep 'em coming if you can!
Well, I still think your leap of logic is bigger than mine.
But anyway, both versions of the question have been answered, so the poster should be happy.
I'm not sure that knowledge implies intent like you suggest.
I could throw a dice and say "Oh look. It's not a 6." But that doesn't mean I meant for it to not be a 6. It's just what happened.
He calls them line segments. For me, that makes things fairly unambiguous.
I think this question is actually a lot easier than that. You're stumbling because the line segments not being parallel doesn't necessarily mean that they'll intersect.
For lines AB and CD to cross, their order around the circle needs to be ACBD or ADBC.
Overall, the possibilities for orders are:
ABCD
ABDC
ACBD
ACDB
ADBC
ADCB
They're all equally likely, so the probability is 2/6 = 1/3.
I haven't yet thought about "a boy born on Tuesday" one...
But for the first one, the only possibilities are...
1. Both children are boys. (√)
2. Both children are girls. (x)
3. One of the children is boy and the other one is girls. (√)I agree that the question is not framed properly since nothing is mentioned about who's younger/elder.
The question is analogous to possessing (order doesn't matter) two balls of different colors (say PINK and BLUE) and not to drawing two balls from two sacs of two different colored balls wherein the order of ball drawn matters!
Yes, those are the only possiblities. But they're not equally likely.
You can try it yourself, by flipping coins. Flip 2 coins at a time and the possible results are:
- Two heads
- Two tails
- One of each
The third result should come up roughly twice as much as the other two.
It's an interesting question though, MIF. I've half got my head around it now.
The two extremes of the question are:
"I have two children. This one here, who I'm directly pointing at with no possiblility for confusion, is a boy. What's the probability that I have two boys?"
Answer 1/2.
"I have two children. One of them, and I'm not giving you any information about which, is a boy.
What's the probability that I have two boys?"
Answer 1/3.
The rest of this type of question form a 'sliding scale'. The probability depends on how likely it is that the given information could describe either child.
So, if he says "One of them likes ice-cream and is a boy", then the answer is close to 1/3, since clearly everyone loves ice-cream.
With the Tuesday clue it's less likely that the two children could be confused, and so the answer is closer to 1/2.
If you said that one child was born on Tuesday while football was on TV and it was a bit cloudy and in the room next door somebody was listening to Lady Gaga, and is a boy, then you're almost certainly specifying the child you mean. So the probability there is 1/2 - ε
I reckon the best policy here is to take a wild stab at the rule and make a list accordingly.
Cracked it. Hopefully.
I don't know if that's the correct term, but it's the one I use.
It's essentially when you think about the fact that a puzzle is a puzzle, rather than its content.
Interestingly, you can take a shortcut with this puzzle by "meta-thinking", making the assumption that there is a single correct answer.
"I was telling my friend about a survey I took about house numbers. I told her how many houses were surveyed, and the sum of the surveyed house numbers. She knew that my number was even, and then she managed to figure out what my number was.
What's my house number?"
Tough:
A bag contains 9 red marbles and 15 blue marbles. Marbles are drawn from the bag without replacement. When the last red marble is withdrawn what is the expected number of blue marbles that have also been withdrawn? Original work please I know my solution and the other one.
Now I've seen the answer, there's probably a far nicer way of doing this. But here's my way:
Nice page!
Two things:
- When you click on the cell of a matrix, I think it'd be good if the current contents got highlighted so you can type over them.
- When a result has a big negative number (big in size, so -0.00001 would count) then trying to transcribe the result over to a matrix makes that cell only display "-". The value is stored properly though, so it's just a visual thing.
This thing is scary.
It asked me 19 questions, and I gave it maybe 4 yesses and 15 nos. It didn't seem to be focussing in on anything in particular. As far as I can tell, the only thing it knew about my character is that he is a British adult male who works on TV.
Then out of nowhere, the right answer comes up.
I really have no idea how that could have happened.
If the width of the room is x, then the length is x+17.
The diagonal of that room is √(x² + (x+17)²).
So the question in the form of an equation is x² + (x+17)² = 54².
(It'll have two answers, but you can ignore the negative one)
I can't find that problem on the page you linked, but I agree with your answer. Given that no bus has come in 20 minutes, the waiting time has to be somewhere in the interval [0,40].