Since the student is guaranteed seven correct responses, she needs at least three of the remaining questions to be correct. Thus we need the probability that she gets 3,4, or 5 of the remaining questions correct.

This situation is an example of a binomial probability distribution. Each event can be described in a binary fashion (either correct or incorrect), each event is independent (the results of one does not effect the others), the probabilities do not change, and there are a finite number of events.

The formula for a binomial probability is given by:

`P(x=k)= ._(n)C_(k)(p)^(k)(1-p)^(n-k)`

where k is the number of successes we seek, n is the number of trials, and p is the probability of success. (For example, suppose we want the probability of 3 successes. The probability of the 1st success is 1/4, the next 1/4, and the 3rd 1/4. The remaining events are failures with a probability of 3/4 each. Using the multiplication principle, we get `(1/4)^3 * (3/4)^2` . However, this can occur in 10 different ways. Suppose we number the 5 questions 1–5. Then she could give correct answers to questions 123, 124, 125, 134, 135, 145, 234, 235, 245, or 345.)

Here n=5, p=1/4, and we want k=3,4, or 5.

`P(x=3)= ._5C_3(1/4)^3(3/4)^2=10(1/64)(9/16)=90/1024`

`P(x=4) = ._5C_4(1/4)^4(3/4)=5(1/256)(3/4)=15/1024 `

`P(x=5) = ._5C_5(1/4)^5(3/4)^0=1/1024`

**Since these are mutually exclusive, the probability of the three events is the sum of the probabilities: P(x=3,4, or 5)=106/1024=53/512 or about 10%.**

**Further Reading**

There are 4 answers for each multiple choice question and each of them has only one correct answer. If any answer is chosen at random the probability of it being correct is 1/4. The probability that it is incorrect is 3/4.

In the 12 question test, the student knows the correct answer to 7 of the questions. For the other 5 questions the answers are picked at random. To get at least 10 answers right, the student should get either 3 or 4 or 5 answers correct for the 5 questions she does not know the answer to.

The probability of this is `(1/4)^3*(3/4)^2` + `(1/4)^4*(3/4)` + `(1/4)^5` = `13/1024`

**The required probability is **`13/1024`