Friday, October 28, 2011

unknown question - random answer - probability via trial

If you choose an answer to this question at random, what is the chance that you will be correct?
a) 25%
b) 50%
c) 60%
d) 25%

Here are my musing - maybe correct - maybe not.

Assume (correctly?) that you do not know enough about the question and answer to know which is the correct answer.

But assume that one of the answers is correct.

There are four possibilities required* - (a) is correct, (b) is correct, (c) is correct, (d) is correct

( *this statement is a little naïve, but a full analysis using methods of probability would I think give you a similar 6/16 conclusion ... although it might produce a more satisfactory explanation )

Count the ticks (6) and the total cells (16) and here is your answer   37.5%   (longhand 6/16)

But does this feel right?

Well if there were just 3 choices (a), (b), (c) and no two answers were the same then you would think 33% (approx)

Note: That assumes that one of the answers is correct - an important assumption.

Preliminary Musings:

Answer is 37.5% providing that the original question is extended to say "...and one of the answers is correct"

Without assuming one of the answers is correct, my reasoning would be different (see below)

But 'Possibility 1' and 'Possibility 4' are the same I hear you cry?
'At random' in this case makes me think 4 trials...

To elaborate:- In my table layout, "(a) is correct" is different to "(d) is correct"

Here I have still ignored the actual text listed in the answers written.

Preliminary Musings 2:

Some reasoning and context: Assuming my answer 37.5% is correct, then going back to the original context ....

...we might say that "...none of the answers is correct" as only options available are 25%, 50%, and 60%

So "what is the chance that you will be correct" could only be answered with NO CHANCE (zero percent)


Looking at the actual answers and reasoning using logic.

25% being the correct answer would mean that (a) is correct answer and (d) is correct answer, but that would indicate 50% success of probability => (a) and (d) are wrong.

(b) being correct would give 25% probability of success, but answer lists 50% => (b) is wrong

(c) being correct would give 25% probability of success, but answer lists 60% => (c) is wrong

Therefore you must conclude that (a) is wrong, (b) is wrong, (c) is wrong, (d) is wrong

So "what is the chance that you will be correct" could only be answered with NO CHANCE (zero percent)

Why is this exercise so hard? Because it is a mixture of probability, logic, abstraction, intuition, counter-intuition, all of which are hard skills.

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