Thats messed up. I voted for A, and it told me I was wrong!
[edit]Probably because I only voted for a single card. oh well.
[/edit]
Marty
[edit]Probably because I only voted for a single card. oh well.
[/edit]Marty
To prove, you must disprove. Therefore, you would need to turn over all four cards in the set to validate your assertion. You would turn over "A" to make sure there is an odd number on the other side; you would turn over "B" to make sure there isn't an odd number on the other side; you would turn over "1" to make sure there is a vowel on the other side; and you would turn over "2" to make sure there isn't a vowel on the other side.
I think I recall saying that I'd tell everyone the answer.
Well, in case anyone hasn't worked it out yet, the answer is A and 2.
Here's why:
You are told that vowels have odd numbers on the reverse. In order to test this you have to be able to disprove the statement. You therefore only need to turn over cards which have the potential to disprove this.
You need to turn over 'A' to test this. If you turn over 'A' and find an even number you have successfully disproved this statement.
You need to turn over '2' to test this as well. If you turn '2' over and find a vowel then this statement is false; vowels can only be paired with odd numbers.
No mention is made of consonants, so there is no need to turn 'B' over, because it will tell you nothing. If you turn it over and find an odd number, it is irrelevant. If you turn it over and find an even number, then it, too, is irrelevant.
Similarly, it is not necessary to turn over '1'. If you turn it over and find a vowel, then the statement is supported. remember, you only need to turn a card over if it can disprove the statement. However, if you turn it over and find a consonant, then the argument becomes that used for 'B'.
Well, in case anyone hasn't worked it out yet, the answer is A and 2.
Here's why:
You are told that vowels have odd numbers on the reverse. In order to test this you have to be able to disprove the statement. You therefore only need to turn over cards which have the potential to disprove this.
You need to turn over 'A' to test this. If you turn over 'A' and find an even number you have successfully disproved this statement.
You need to turn over '2' to test this as well. If you turn '2' over and find a vowel then this statement is false; vowels can only be paired with odd numbers.
No mention is made of consonants, so there is no need to turn 'B' over, because it will tell you nothing. If you turn it over and find an odd number, it is irrelevant. If you turn it over and find an even number, then it, too, is irrelevant.
Similarly, it is not necessary to turn over '1'. If you turn it over and find a vowel, then the statement is supported. remember, you only need to turn a card over if it can disprove the statement. However, if you turn it over and find a consonant, then the argument becomes that used for 'B'.
mark your solution depends on your interpritation of "always". Let me explain.
Way 1: Always means that it, and only it, can be true.
Example: John always gets sprinkles on his Vanilla ice cream, but he likes strawberry flavor too. In that case he can get sprinkles with vanilla and ONLY vanilla - sprinkles and strawberry is impossible.
Way 2: Always means that in that case its always true, but others are not ruled out.
Example: John always gets sprinkles on his Vanilla ice cream, but he likes strawberry flavor too. In this case whenever he gets Vanilla ice cream he will ALWAYS, without failure, get sprinkles, but if he then decides to get strawberry flavored ice cream he could get sprinkles on that too (he doesnt have to, but he could if he wants them).
I hope this makes sense to everyone, because the answer to the question really does depend on your interpritation of the word "always" in the problem.
Way 1: Always means that it, and only it, can be true.
Example: John always gets sprinkles on his Vanilla ice cream, but he likes strawberry flavor too. In that case he can get sprinkles with vanilla and ONLY vanilla - sprinkles and strawberry is impossible.
Way 2: Always means that in that case its always true, but others are not ruled out.
Example: John always gets sprinkles on his Vanilla ice cream, but he likes strawberry flavor too. In this case whenever he gets Vanilla ice cream he will ALWAYS, without failure, get sprinkles, but if he then decides to get strawberry flavored ice cream he could get sprinkles on that too (he doesnt have to, but he could if he wants them).
I hope this makes sense to everyone, because the answer to the question really does depend on your interpritation of the word "always" in the problem.
Hmm. OK, I see your point.
That rather b****rs my experiment. I had hoped to prove that people were not logical and would usually make the mistake of choosing A and 1.
Anyway, here are my statistics (after removal of duplicates - yes some people really were boring enough to delete the cookie and resubmit the same answer 90 times):
Answer Count Of Answers
1 (2) 4
2 (1) 26
3 (1+2) 7
4 (B) 11
5 (B+2) 21
6 (B+1) 11
7 (B+1+2) 1
8 (A) 34
9 (A+2) 56
10 (A+1) 67
11 (A+1+2) 6
12 (A+B) 6
13 (A+B+2) 1
14 (A+B+1) 3
15 (A+B+1+2) 28
As I expected more people chose A+1 than A+2, with relatively fewer people opting for B, either on its own, or with others.
Might have to think up another experiment like this, but maybe I should check my wording more thoroughly
That rather b****rs my experiment. I had hoped to prove that people were not logical and would usually make the mistake of choosing A and 1.
Anyway, here are my statistics (after removal of duplicates - yes some people really were boring enough to delete the cookie and resubmit the same answer 90 times):
Answer Count Of Answers
1 (2) 4
2 (1) 26
3 (1+2) 7
4 (B) 11
5 (B+2) 21
6 (B+1) 11
7 (B+1+2) 1
8 (A) 34
9 (A+2) 56
10 (A+1) 67
11 (A+1+2) 6
12 (A+B) 6
13 (A+B+2) 1
14 (A+B+1) 3
15 (A+B+1+2) 28
As I expected more people chose A+1 than A+2, with relatively fewer people opting for B, either on its own, or with others.
Might have to think up another experiment like this, but maybe I should check my wording more thoroughly
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