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April 2nd, 2012, 12:40 AM  #1 
Newbie Joined: Apr 2012 Posts: 7 Thanks: 0  Confusing probability question
A student is taking a multiplechoice exam in which each question has 4 choices. Assuming that she has no knowledge of the correct answers to any of the questions, she has decided on a strategy in which she will place 4 balls (marked A,B,C,D) into a box. She randomly selects 1 ball for each question and replaces the ball in the box. The marking on the ball will determine her answer to the question. There are 4 multiple choice questions on the exam. (a) What is the probability that she will get 4 questions correct? (b) At least 3 questions correct? (c) No more than 2 questions correct? My teacher left early since he had to go somewhere, so he didn't explain to us how to do this question. Again, much help would be very appreciated! 
April 2nd, 2012, 05:39 AM  #2 
Newbie Joined: Apr 2012 Posts: 7 Thanks: 0  Re: Confusing probability question
Attempted to work these out. But my brain is all clustered atm.. I really need to work on my probability/mathematics skills. (a) If it's a 1/4 chance of getting one right for each qq. 4 qq's = 16 possible chances, so 4/16 to get 4 qq's correct. (25%)? (b) Confused with the wording [at least 3 qq's correct] (c) Confused with the wording for this as well [no more than 2 qq's correct] 
April 2nd, 2012, 05:44 AM  #3  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond  Re: Confusing probability question Quote:
 
April 2nd, 2012, 06:40 AM  #4  
Senior Member Joined: Dec 2011 Posts: 277 Thanks: 2  Re: Confusing probability question Quote:
Different people have different problemsolving style preferences. Well, I'll just follow [color=#00BF00]greg1313[/color]'s approach. P(At least 3 questions correct) =P(3 questions correct or 4 questions correct) =P(3 questions correct) or P(4 questions correct) =P(3 questions correct and 1 wrong) or P(4 questions correct) + P(CCCC)" />  

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