SOLUTION: I am trying to obtain the probability of randomly answering a multiple choice question (select all that apply) correctly, depending on whether selecting 1, 2, 3, 4 or all 5 options
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Question 714642: I am trying to obtain the probability of randomly answering a multiple choice question (select all that apply) correctly, depending on whether selecting 1, 2, 3, 4 or all 5 options of the question options is correct. For example, a multiple choice questions (select all that apply) has 5 multiple choices of which choices "A", "D" and "E" are correct and the student would not get credit for selecting any other combination of the five choices or subset of the correct choices {"A", "D" and "E"} (i.e., selects "A" and "E", only). I have tried summing all of the r-combinations of S={A, B, C, D, E}:
5!/r!(5-r)!, for r=1 to r=5; the answer which I get is 31. I therefore think that the probability of randomly selecting "A", "D" and "E" is 1/31 = 0.03226.
Is this correct? I would appreciate your help,
Lucas
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
From the way you worded the question, I would presume that selecting none of the five choices is NOT an option, i.e., every random selection would include at least one of the choices "A" through "E". In this case, your answer of 1/31 is spot on. On the other hand, if not making any selection is a valid option for answering the question, the correct probability would be 1/32 because you would have to calculate:
instead of
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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