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This only produces a paradox if you fall for the usual fallacy that "at random" necessarily means "with uniform probability".
For example, I would pick an answer at random by rolling a fair cubic die and picking a) if it rolls a 1, b) on a 2, d) on a 3 or c) otherwise so for me the answer is c) 50%.
However, as it specifies that you are to pick at random the existence, uniqueness and value of the correct answer depends on the specific distribution you choose.
It's annoying that 25% appears twice. How about these answers:
a) 100%
b) 75%
c) 50%
d) 0%
Since two of them are the same, you have a 50% chance of picking something that is 33% of the possible answers. The other two, you have 25% chance of picking something that us 33% of the possible answers.
So 50%33% + 2 (33%*25%)= 33%
So your chances of being right is 33% cause there is effectively 3 choices.
But that one answer has a 33% larger possibility of being chosen by random, than the remaining two.
I covered that by multiplying it by 50% as it represents 50% of the choices.
100 **** percent, i'm all in!
I argue it's still 25%, because the answer is either a,b,c, or d, you can only choose 1, regardless of the possible answer having two slots.
Yup. And it says pick at random. Not apply a bunch of bullshit self mastubatory lines of thinking. Ultimately, 1 of those answers are keyed as correct, 3 are not. It's 25% if you pick at random. If you're applying a bunch of logic into it you're no longer following the parameters anyway.
You can just say "I don't understand probability (or the word 'if')" next time and save a whole bunch of effort.
Ahhhhhhhhhhhhhhhhh
There's a reason I dropped probability at school.
There is absolutely no way it is 60%. Because you can never have 60% chances of picking anything particular when there are only 4 choices. Knowing this, the answer is either 25% or 50%. Two effective choices, so the answer is C, 50%.
60%
B) 60% because I'm generally very lucky.
Can I take a 50/50 joker first?