Hi Debbie. I'll show you with a similar test.

What is the probability of getting 2 out of 4 (a, b, c, d) questions correct by random selection?

It could be any set of two out of the four questions, so we start with ₄C₂ (that's _questionsC_correct) possible choices.

For each of those, we know that each of the questions we got wrong had three possible choices of "wrong" answer. So there are 3^(4-2) (that's [wrong choices]^{(wrong questions)} ) choices for wrong answers.

That means that there are ₄C₂ × 3² ways of getting 2 out of 4 on this test.

Now how many ways are there of answering the test in all? Since each question has 4 choices and there are 4 questions, there are 4⁴ (that's choices^questions) possible ways of answering the test.

Thus, the probability is (the number of ways we can get 2 out of 4) divided by the (number of possible ways of answering in total).

So the probability for my situation is ( ₄C₂ × 3² ) / 4⁴ = (6 × 9) / 256 = 54 / 256 = 27 / 128. You can answer your problem the same way.

Here's the proof that my answer is correct: I will write out all the possibilities.

Now you try it with your problem.

Cheers,
Stephen La Rocque