Prob. of passing multiple choice test, w/ & w/o eliminating

A student is taking a test consisting of multiple choice questions, each one with four options.

1. If a student does not eliminate any options and guesses randomly on all questions, what is the probability that he will achieve a score of 70% or higher?

2. If a student eliminates two out of the four options for each question and chooses randomly from the remaining options, what is the probability that he will achieve a score of 70% or higher?

If possible, can you show me the equation or function you use to actually solve this problem, so that I can solve it while changing one or more of the variables?

Thanks!

Re: Prob. of passing multiple choice test, w/ & w/o eliminat

One Question

No Elimination
p(>70%) = p(correct) = 0.25

Eliminate 2
p(>70%) = p(correct) = 0.50

Okay, now you do 2 questions.

Re: Prob. of passing multiple choice test, w/ & w/o eliminat

I don't follow...

I know that the probability without eliminating, on any single question, would be 0.25. And of course, it stands to reason that the probability if two options are eliminated is double that, so 0.50.

That would seem to suggest that the probability of getting 25% of the questions correct is 1.0. How is the probability of getting 70% or more 0.25?

I don't see how you got either of your answers.

Also, I don't know what you're asking me to do next.

Re: Prob. of passing multiple choice test, w/ & w/o eliminat

There we have something to work with.

If it "stands to reason", what happens when we apply this method to four (4) questions?

1 question ==> 0.25
2 questions ==> 0.50
3 questions ==> ??
4 questions ==> ??

What is the implication of your result for 4 questions?

Note: Getting one question correct does not imply you get the second question correct.