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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 multiple-choice 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
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Re: Confusing probability question

Quote:
 Originally Posted by HungerGames1 (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%)?
1/4 * 1/4 * 1/4 * 1/4 = 1/256.

April 2nd, 2012, 06:40 AM   #4
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Re: Confusing probability question

Quote:
 Originally Posted by HungerGames1 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]
b.
Different people have different problem-solving 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' + P(CCCC)" />

$=\frac{1}{4}.\frac{1}{4}.\frac{1}{4}.\frac{3}{4}.\ frac{4!}{3!} +\frac{1}{4}.\frac{1}{4}.\frac{1}{4}.\frac{1}{4}$

$=\frac{3}{64}+\frac{1}{256}$

$=\frac{13}{256}$

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# A student is appearing in a multiple-choice exam in which each question has four 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 four balls (marked A, B, C

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