Solved on Sep 18, 2023

Exam with 7 multiple-choice questions, 3 answers each. Find the number of possible answer combinations and the probabilities of a student randomly guessing all correct, all incorrect, or 2 correct and the rest incorrect.
1) How many ways can the exam be answered? 373^7 2) Probability of all answers correct: (13)7\left(\frac{1}{3}\right)^7 or 0.0000%0.0000\% 3) Probability of all answers incorrect: (23)7\left(\frac{2}{3}\right)^7 or 0.4629%0.4629\% 4) Probability of 2 correct (questions 5 and 6) and the rest incorrect: (13)2(23)5\left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^5 or 0.0123%0.0123\%

STEP 1

Assumptions1. There are7 questions in the exam. . Each question has3 possible answers, with exactly one correct answer.
3. The student is guessing randomly, so each answer has an equal chance of being chosen.

STEP 2

First, we need to find the total number of different ways to answer all the questions on the exam. We can do this by using the Fundamental Counting Principle, which states that if there are n ways to do one thing and m ways to do another, then there are n*m ways to do both.
Totalways=NumberofchoicesNumberofquestionsTotal\, ways = Number\, of\, choices^{Number\, of\, questions}

STEP 3

Now, plug in the given values for the number of choices and the number of questions to calculate the total ways.
Totalways=37Total\, ways =3^{7}

STEP 4

Calculate the total number of ways to answer all the questions.
Totalways=37=2187Total\, ways =3^{7} =2187

STEP 5

Now, we can calculate the probability of the student getting all the answers correct. Since there is only one correct answer for each question, the number of ways to get all the answers correct is1^7.
Probabilityofallcorrect=NumberofwaystogetallcorrectTotalwaysProbability\, of\, all\, correct = \frac{Number\, of\, ways\, to\, get\, all\, correct}{Total\, ways}

STEP 6

Plug in the values for the number of ways to get all correct and the total ways to calculate the probability.
Probabilityofallcorrect=1218Probability\, of\, all\, correct = \frac{1^{}}{218}

STEP 7

Calculate the probability of getting all answers correct.
Probabilityofallcorrect=172187=12187Probability\, of\, all\, correct = \frac{1^{7}}{2187} = \frac{1}{2187}

STEP 8

Convert the probability to a percentage and round to four decimal places.
Probabilityofallcorrect=12187×100%=0.0457%Probability\, of\, all\, correct = \frac{1}{2187} \times100\% =0.0457\%

STEP 9

Now, we can calculate the probability of the student getting all the answers wrong. Since there are2 wrong answers for each question, the number of ways to get all the answers wrong is2^7.
Probabilityofallwrong=NumberofwaystogetallwrongTotalwaysProbability\, of\, all\, wrong = \frac{Number\, of\, ways\, to\, get\, all\, wrong}{Total\, ways}

STEP 10

Plug in the values for the number of ways to get all wrong and the total ways to calculate the probability.
Probabilityofallwrong=272187Probability\, of\, all\, wrong = \frac{2^{7}}{2187}

STEP 11

Calculate the probability of getting all answers wrong.
Probabilityofallwrong=72187=1282187Probability\, of\, all\, wrong = \frac{^{7}}{2187} = \frac{128}{2187}

STEP 12

Convert the probability to a percentage and round to four decimal places.
Probabilityofallwrong=1282187×100%=5.8577%Probability\, of\, all\, wrong = \frac{128}{2187} \times100\% =5.8577\%

STEP 13

Finally, we can calculate the probability of the student getting questions5 and6 correct and the rest of the answers wrong. There is way to get each of these questions correct and2 ways to get each of the other questions wrong.
Probabilityof5and6correctandrestwrong=Numberofwaystoget5and6correctandrestwrongTotalwaysProbability\, of\,5\, and\,6\, correct\, and\, rest\, wrong = \frac{Number\, of\, ways\, to\, get\,5\, and\,6\, correct\, and\, rest\, wrong}{Total\, ways}

STEP 14

Plug in the values for the number of ways to get and6 correct and the rest wrong and the total ways to calculate the probability.
Probabilityofand6correctandrestwrong=22187Probability\, of\,\, and\,6\, correct\, and\, rest\, wrong = \frac{2^{}}{2187}

STEP 15

Calculate the probability of getting questions5 and correct and the rest of the answers wrong.
Probabilityof5andcorrectandrestwrong=252187=322187Probability\, of\,5\, and\,\, correct\, and\, rest\, wrong = \frac{2^{5}}{2187} = \frac{32}{2187}

STEP 16

Convert the probability to a percentage and round to four decimal places.
Probabilityof5and6correctandrestwrong=32218×100%=.4634%Probability\, of\,5\, and\,6\, correct\, and\, rest\, wrong = \frac{32}{218} \times100\% =.4634\%a) The probability the student got all of the answers correct is0.045%. b) The probability the student got all of the answers wrong is5.857%. c) The probability the student got questions5 and6 correct and the rest of the answers wrong is.4634%.

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