the question is #20 on Chapter 5: A surprise quiz contains three multiple-choice

Question

the question is #20 on Chapter 5: A surprise quiz contains three multiple-choice questions: Q1has four suggested answers, Q2 has three, Q3 has two. A completelyunrprepared student decides to choose the answwers at random. Let Xdenote the number of questions the student answers correctly. so, i know the X = {0,1,2,3,4) Yet i'm not sure about the rest of the part, the probabilityof distributions of X. There are 24 different kinds of ways b/c 4*3*2*1=24. 0 correct would be 3/4*2/3*1/2=1/4 3 correct would be 1/4*1/3*1/2=1/24 Then what about 1 correct and 2 correct? plz, help me out. the question is #20 on Chapter 5: A surprise quiz contains three multiple-choice questions: Q1has four suggested answers, Q2 has three, Q3 has two. A completelyunrprepared student decides to choose the answwers at random. Let Xdenote the number of questions the student answers correctly. so, i know the X = {0,1,2,3,4) Yet i'm not sure about the rest of the part, the probabilityof distributions of X. There are 24 different kinds of ways b/c 4*3*2*1=24. 0 correct would be 3/4*2/3*1/2=1/4 3 correct would be 1/4*1/3*1/2=1/24 Then what about 1 correct and 2 correct? plz, help me out.

Explanation / Answer

Ok, first off  X = {0,1,2,3} there can't be a 4because there's only 3 questions. Probability distribution: 2 correct: the possible combinations of two correctare: Q1 & Q2, Q1 & Q3, and Q2 & Q3 =1/4*1/3*1/2 + 1/4*2/3*1/2 + 3/4*1/3*1/2 = 1/24 + 2/24 + 3/24 = 6/24= 1/4 1 correct is either Q1, Q2, or Q3= 1/4*2/3*1/2 +3/4*1/3*1/2 + 3/4*2/3*1/2 = 2/24 + 3/24 + 6/24 = 11/24 Then to double check that they all add up to one: 6/24 + 11/24+ 6/24 + 1/24 = 24/24 =1

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