If you choose the answer to this question at random, what is the probability that it will be correct?
a) \(25 \) %
b) \(50\) %
c) \(0\) %
d) \(25\) %
Good Luck. :)
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Muzaffar Ahmed
3 years, 10 months ago
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Top Newest25%
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25 % . Because there are four options each contributing 25 %
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Not so simple, man, not so simple
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pray tell us the answer!!!!!!!!!
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I guess these sort of questions are good for discussions rather for finding answers. For any answer there is a contradictory statement ......:) how about considering both 25% into single unit and the remaining 0 and 50 we get one more combination 1/3*100
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They are not one unit.. They are different options and the question is about choosing one option randomly...
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50%
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Why 50% ?
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25%
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Why 25% ?
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50%
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0% is the correct answer .............. if it would have been 25% then if we chose randomly, there are 2 choices out of a total of four , that seem optimal ( i.e. option(a and (d ) so that would mean that the probability would be equal to 50 % ........ Contradictory! if it would have been 50% , then it would have meant that only one choice out of the four is correct i.e. 25%... this answer is also contradictory to itself.......... if it would have been 0% then again it would imply that one choice out of the four is correct i.e. 25% ...... hence it seems that none of the option is correct ......... that leads us to the answer 0% , and hence this answer is not in contrary to itself!!!!!!!
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If 0% is correct, then it is one of the four options, which would give you 25% probability again.
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Agree. Can't be zero percent.If one of the options is correct and the user selects it randomly it definitely will have a probability. I will still stick to 25% . because among 4 options only one is correct and the probability for that becomes 25%(1/4 multiloed by 100)
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There are 2 instances of the option 25%, so if 25% is correct, you will have \( \frac{2}{4} \times 100 = 50\) % probability.
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Oop missed it didn't notice 25 repeating again
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