Question 316451: if you make guesses for four multiple choice test questions (each with 5 possible answers), what is the probability of getting at least one correct?
Answer by moshiz08(60)   (Show Source):
You can put this solution on YOUR website! Let's find the probability that you get all of them wrong.
On the first question, there are 4 wrong answers out of 5. So the probability that you are wrong is 80% or 0.8
Same for the second, third, and fourth questions: you have 80% chance of getting it wrong.
The probability that you get all wrong is .8 * .8 * .8 * .8. (We assume that the events are independent so we can multiply their probabilities. For example, what is the probability that if you flip a coin two times you will get all heads? You can make a list of possibilities HH, HT, TH, TT, so it is 1 out of 4 or 25%. Alternatively, the probability of the first one being heads is 50% and the probability of the second one being heads is also 50% so the probability of both being heads is .5 * .5 = .25 or 25%.)
Anyway, the probability of getting all wrong is thus .4096 or 40.96%.
But we wanted to know what is the probability of getting at least one right, that is the same as NOT getting all wrong. If the probability of getting all wrong is 40.96%, then the probability of NOT getting them all wrong is 59.04%.
Think about it like this: you can get all of them wrong or you can get at least one right. These are the only events possible - it has to be one or the other, nothing else. Thus, their probabilities must add to 100%. In general, for an event X with a probability P, the probability that X does NOT happen is 1 - P.
Now we could have found the probability that you get one question right, plus the probability that you get 2 right, plus the probability that you get 3 right, plus the probability that you get all 4 right, but this is a shortcut.
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