A multiple choice test consists of 5 questions each with 4possible answers of wh

Question

A multiple choice test consists of 5 questions each with 4possible answers of which only one si correct. a) in how many ways can a student answer the test? b) in how many ways can a student get no correct answer? c) in how many ways can a student get all correctanswers? d) in how many ways can a student get exactly 2 correctanswers? A multiple choice test consists of 5 questions each with 4possible answers of which only one si correct. a) in how many ways can a student answer the test? b) in how many ways can a student get no correct answer? c) in how many ways can a student get all correctanswers? d) in how many ways can a student get exactly 2 correctanswers?

Explanation / Answer

The first part is more straightforward than (b) or (d). How many ways can a student answer the test? Well, to startwith, she has 4 possibilities for the first question. Then,for each of those possibilities, she moves on to having 4possibilities for the second question. So, after twoquestions, there are 16 possible ways to answerthe test:  aa, ab, ac, ad, ba, bb, bc, bd, ca, cb,cc, cd, da, db, dc, and dd. Then, for whichever one of thesecombinations of answers she picks, there are another 4 ways toanswer the third question. So if there were three questionson the test, there would be 4^3 = 4*4*4 = 64 possible ways toanswer the test. I'll let you extrapolate to 5questions. . Part (b) is very similar to part a, except that now, for eachquestion, there are only 3 possible answers, because we're lookingat only the wrong answers. So if it were a three questiontest, there would be 3^3 = 27 ways to miss all of them. . Part (c) is hopefully something you can do on your own. . How many ways can a student get exactly 2 correct answers isthe trickiest part. Let's do this in two parts:     Part 1: identify all the ways that thestudent could get exactly 2 correct answers, if we know which 2questions she got right, and then     Part 2: identify all the differentcombinations of which questions she gets right . Part 1: suppose we knew she got the first 2 correct, and thenmissed the last 3. For the first 2 questions, there'd beexactly 1 way to get each one right, so so far we have only 1way. Now we need to consider all the ways to miss the last 3questions. For each of the last three questions, there are 3ways to get the question wrong. So how many ways are there toget the first two questions right and miss the last 3? . Part 2: Now we consider the problem of not knowing which 2questions she got right. In part 1, we assumed that she gotthe first 2 right (RRWWW), but it could have been WWWRR, or RWRWW,or WWRRW, or any of many other possible combinations. Howmany combinations are there? . This is a combinatorics problem, where we have to choose TWOpossibilities (the two questions to get right) out of FIVEpossibilities.  Hopefully you remember the formula forpicking k options out of n possibilities: n! / [k!(n-k!)] . So if we were choosing 4 questions to get right out of 6,the total number of combinations would be 6C4 = 6!(4!2!) =6*5*4! / (4!*2) = 6*5/2 = 3*5 = 15. (Hint: when doingcombinatorics, never start by multiplying out all the factorials inthe numerator and denominator. You can cancel a heck of a lotbefore you start multiplying.) . So how many ways can you choose 2 questions to get right outof 5 questions? For each of those combinations, how many waysare there to miss the questions that you missed? Multiplythose two numbers together, and you'll have your answer.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.