Multiple Choice Tests
Date: 05/06/97 at 02:41:28 From: nancy Subject: Probability I have a question on my math homework that I can't figure out. I think it deals with Pascal's triangle, but I just don't understand it. If you have a test with 6 questions and 4 possible answers to each question, and you guess all answers, what is the probability of getting all 6 questions right? I tried using the 6th row and the 4th column of Pascal's triangle, but I don't think that is right. Can you please help me?
Date: 05/06/97 at 05:53:28 From: Doctor Anthony Subject: Re: probability This is relatively easy because you are asked for the probability that you get all 6 questions right. The probability of getting any one question right is 1/4 because there are 4 possible answers, only one of which is correct. The probability of getting all six questions right is the probability of getting one question right times itself six times: 1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 = (1/4)^6 = 1/4096 Not very good odds! It's better to study so you don't have to guess! A more difficult problem would be to figure out the probability of getting 3 out of the 6 correct. Suppose you got the first three correct and the last three wrong. We already know the probability of getting one question right: 1/4. This means the probability of getting one question wrong must be 3/4 (you have to either get it right or wrong - there's no middle ground here). The probability of this particular sequence of right and wrong answers would be: 1/4 x 1/4 x 1/4 x 3/4 x 3/4 x 3/4 = (1/4)^3 (3/4)^3 However there are many possible sequences for right and wrong answers giving 3 correct and 3 incorrect. This is where the terms of Pascal's triangle would come in. The number of possible sequences in this case is given by: 6.5.4 6_C_3 = ------- = 20 1.2.3 So there are 20 possible sequences giving three correct and three incorrect answers. The probability of exactly three correct answers is: 20(1/4)^3 (3/4)^3 = 540/4096 = 135/1024 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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