Another Maths Problem For You

[FrogNog]

If you choose an answer to this question at random, what is the chance you will be correct?

a) 25%

b) 50%

c) 60%

d) 25%


Over to you.

[bhg481]

No, OG, not an invalid question. The question is "what is the chance you will be correct?". As you said, none of them can be correct, so the answer to the question is 0%.

[Old_Geezer]

And since it is stated that you have to chose one of the options as your actual answer, and none of them are 0%, that is what makes it invalid.

[bhg481]

No, it doesn't say you have to. It says IF you choose one, emphasis IF. IF you choose one you will always be wrong.

[Old_Geezer]

You need to read through the follow up posts, that is where it is clarified.

[bhg481]

On the contrary, the OP states the question, follow-up posts are opinions.

[jim360]

It would depend on how the question was presented. Since it's never going to be asked in a serious exam, of course, we'll never know, but there are two options: either you are obliged to choose one of a, b, c, d as your answer, in which case there is no valid answer, or you are able to provide a different answer from those presented, in which case you could write down 0% and be correct.

Usually it's implicit in multiple-choice questions that the answer has to be one of those given. On occasion this has hurt students taking real exams because it turns out that there has been some error along the line, and the poor student has to circle the correct answer to "What is 2+1? Is it a) 4, b 5 or c) 7?" and will of course find that he can't circle anything and be right. The sad thing about exams is that implicitly you're trusting the examiner to have checked things, and so inevitably people will tear their hair out trying to work out why 2+1 can't be 3...

Anyway, the answer is either 0% if you are free not to pick the given options, or there is no answer if you have to pick a, b, c or d; or one of the presented options could be correct if you choose a random distribution that ensures that you pick, say, option b) with probability 50% and options a, c or d with total probability 50%.

[shoota]

Sandy - ///50%. Random is the operative word. ///

I agree, but would argue that there are four different options as the two 25% are qualified by the addition of letters, (a & d) so if you were to write these numbers on marbles, put them in a bag and choose one at random by putting your hand in and picking one out unsighted then you would have as much chance of picking a) 25% as d) 25%