How Do You Calculate Probabilities for Guessing on a Multiple Choice Test?

[s.mack77]

Homework Statement

A ten question multiple choice exam is given, and each question has 5 possible answers. Pascal Gonyo takes this exam and guesses at every question. Find the probability that he will get at least 9 questions correct. Without using the binomial probability formula, determine the probability that he gets exactly two questions correct.

Homework Equations

P(at least 9) + P(less than 9) = 1

The Attempt at a Solution

P = 1-P(8)
I'm stuck on this. I found the probability of at least one question correct using this formula but I can't get it to work for 9.

 

[christianjb]

Use the binomial formula for the first part of the question.

 

[Dick]

The probability he will get exactly two questions correct is pretty easy. What is it? As is the probability he will get at least 9 correct is the sum of the probabilities of getting getting exactly 9 correct plus the probability of getting exactly 10 correct. What are they? I'm not sure I see a way to do these without using at least a vestige of the binomial theorem. But someone may prove me wrong tomorrow.

 

[s.mack77]

The binomial formula will give me exactly 9 right? how do I get at least 9?

 

[christianjb]

Add the prob of getting exactly 9 and the prob of getting exactly 10.

 

[s.mack77]

Thanks, can you tell me how to find the probability of getting 2 correct without the binomial formula?

 

[christianjb]

It's a dumb question! Why on Earth should they tell you not to use a particular theorem?

 

[christianjb]

OK, so how many distinct possibilities are there for scoring?

e.g. for 3 Q's there's 8 possibilities

YYY
YYN
YNY
YNN
NYY
NYN
NNY
NNN

For 10 Q's how many distinct possibilities are there?

Oops- there's 5 possible answers per Q. Well- can you work it out for that case?

 

[threetheoreom]

It's a dumb question!

I am quite sure the OP is just looking for some clarity, don't be that harsh.

 

[HallsofIvy]

It's a dumb question! Why on Earth should they tell you not to use a particular theorem?

Perhaps because they want people to think rather than just apply formulas? It wouldn't make sense to work it out from first principles for, say, 4 or 5 out of 10, but for "exactly 2" it shouldn't be too much to expect.

What is the probability of "CCIIIIIIII" where C= correct and I= incorrect?
Can you show that the probability of 2 correct and 8 incorrect is exactly the same no matter what order?
How many different orders are there- in how many ways can you write 2 Cs and 8 Is?

 

[christianjb]

Good grief!

Threetheoreem: I was criticizing the Q, not the poster. The poster presumably didn't compose the question.

HOI: I said it's a dumb question because I don't like the idea of restricting how a student solves a problem. Give me the Q- but don't tell me how to find the A. It also takes intelligence to see if the binomial theorem is applicable in this case.