What is the probability of getting exactly 4 correct answers?

Question

about a multiple-choice test in which each question has 4 choices only one of which is correct. Assume that 9 questions are answered by guessing randomly.
What is the probability of getting exactly 4 correct answers?

Answer 1

Katie,

 

If we just look at a single question, I either get it right, or get it wrong.  The probabilities aren't the same for those two things:  I have a 1 in 4 chance of getting it right, and a 3 in 4 chance of getting it wrong.  But, regardless, it's either right or wrong.

 

What makes these problems a bit messy is that we're dealing with a 9-question test.  If this was about a single question, it's not too horrible.  But here, it's asking the probability of getting four questions right out of nine, so I've got a bit of an issue.  It doesn't tell me which four questions I got right or wrong!  I could answer the first four questions right, or questions 2/5/7/8, or questions 3/6/8/9...you get the idea.  There are a lot of different ways of doing that, and then for each of those questions, I have a 1 in 4 chance that I got it right.  How do we deal with both of those issues at the same time?

 

Thankfully, there's a formula for this.  :)  Because each question on the test is either right or wrong, we're dealing with something called a binomial experiment...and then there's a formula for binomial probabilities. 

 

The probability of getting exactly k successes in n attempts is equal to:

 

P(k successes in n attempts) = _nC_k * (p^k) * (q^n-k)

 

In this specific problem, I've got 9 attempts to get exactly 4 questions right.  So, can you figure out what n and k are from that? 

 

- The big C is "n choose k," which you hopefully have covered in your class...and your calculator will do it for you.

- The little p is "the probability of success on a single attempt."  So, in this problem, what is my chance of getting a single question right?  And then we raise that to the k power.

- The little q is "the probability of failure on a single attempt."  In this problem, what is my chance of getting a single question wrong?

- The "n - k" in the exponent is something we should be able to figure out by now, because we know n and k.  "n - k" will tell us how many failures we'll get out of our n attempts.  (In this problem, how many questions will we get wrong on the test if we got exactly 4 right?)

 

Hope this helps!

 

 

Answer 2

Please see the answer from Michael