# A multiple choice test contains 12 questions, each offering 4 possible answers, only 1 of which is correct. A student knows the correct answer to 7 of the questions but randomly guesses the answers to the remaining 5. What is the probability the student will get at least 10 of the 12 correct? 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%. 