# Thread: Are You Able To Pass The Cambridge Test??

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1. ## Are You Able To Pass The Cambridge Test??

I already know that I am not qualified to enter Cambridge; but I took the test anyway just to see if I got anything right. My instincts were correct as I only got 2 of the 10 correct!!
Give it a shot:
Could you pass a Cambridge University interview? - Telegraph

2. I scored 60%, but would like to see their explanation of the birthday and dice questions. I think they are wrong on those two.

3. 70% here. Although one question was a 50/50 punt that I got right...

For the dice question, as the dice is completely unbiased there is a 1 in 6 chance that you will roll a 6 which means on average a 6 will come up in 6 rolls. This does not mean that a 6 will come up in every group of 6 rolls or that only 1 will. The more rolls you make then the closer it will get to the answer given.

I got the birthday question wrong and am currently trying to understand the answer!

4. \$spacer_open \$spacer_close
5. Nah, I can't answer questions like that. Besides, I failed a real, live Cambridge interview so I know I don't have what it takes.

6. Thank you for completing this quiz.
You scored 30%!
Makes ya wonder how I made it as a moderator and Linux team member on a couple of distros.
I bet the test maker could not rebuild a motorcycle transmission or contruct anything as fine as
I do with nothing avaiable though.

They hurt my feelers with that.

7. Originally Posted by elija
70% here. Although one question was a 50/50 punt that I got right...

For the dice question, as the dice is completely unbiased there is a 1 in 6 chance that you will roll a 6 which means on average a 6 will come up in 6 rolls. This does not mean that a 6 will come up in every group of 6 rolls or that only 1 will. The more rolls you make then the closer it will get to the answer given.

I got the birthday question wrong and am currently trying to understand the answer!
With a 1 in 6 chance, each number is equally likely to come up, so with a large number of rolls, a 6 should come up, on average, in 3.5 rolls out of 6, but the interval between 6's would be 6. I was thinking average rolls to get a 6, not interval. Thanks for getting me on the right track elija.

8. I've looked at two of those questions now, and they've both been incorrectly described or incorrectly worked out.

This one:

An unbiased uniform random number generator, which generates numbers between 0 and 1 inclusive, is used to generate two numbers A and B. Find the probability that A will be greater than B.
suggests the answer is 1/2. This is not true. They claim the answer is:

P( A > B) = P( A < B )

therefore P (A>B) = 1/2.

The correct answer is, of course:

P (A > B) - P (A = B).

Which is very close to half, but not actually a half.

Two particles A and B each of mass m are connected by a light inextensible string with A on a smooth table and B hanging down the string below the table. In order to keep B in equilibrium, what should be the speed of A?
I think that the assumption here is that the particle A is spinning about the hole through which the string hangs - but it doesn't say that. The only way that particle B is in equilibrium, i.e. not going up or down, is for the string to be stationary. If A isn't spinning around the hole (and the question doesn't say that it is) then it's speed MUST be zero, otherwise particle B will be moving in the same speed away from or towards the hole as particle A is. Hence A must be glued to the table. That solves the problem, but I don't think it's what they meant in their badly described question.

If you ask me, those maths professors at Cambridge have some clever ideas about how to test your reasoning, but don't communicate very well.

9. Got lucky on 2 I guessed.

Thank you for completing this quiz.

You scored 100%!
For the Birthday question, think of how many different PAIRS of people are in a group of 23.

10. Got 50%, but I still have not got my morning coffee ...

11. Originally Posted by slw210
Got lucky on 2 I guessed.

For the Birthday question, think of how many different PAIRS of people are in a group of 23.
I've not looked at that question, but I suspect that the boffins at Cambridge got it wrong...

The correct answer is 22+21+20+19+18+17+16+15+14+13+12+11+10+9+8+7+6+5+4 +3+2+1, and I believe this is expressed as 22!

Doesn't this equate to 253?

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