# binomial probability distribution question #4 A student is taking a multiple choice exam in which each question has 4 choices. Assuming that she has no knowledge of the correct answers to any of...

**binomial probability distribution question #4**

A student is taking a multiple choice exam in which each question has 4 choices. Assuming that she has no knowledge of the correct answers to any of the questions, she has decided on a strategy in which she will place 4 balls (marked A, B, C and D) into a box. She randomly selects one ball for each question and replaces the ball in the box. The marking on the ball will determine her answer to the question. There are 5 multiple choice questions on the exam. What is the probability that she will get 5 questions correct?

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### 2 Answers

You need to use binomial probability formula to determine what is the probability that the student to get 5 questions out of 5 correct, hence:

`P(r) = C_n^r*p^r*q^(n-r)`

Notice that the 5 questions are 4 choice questions, hence the probability to get one correct choice out of 4 is: `p = 1/4 = 0.25` .

You need to remember that `p+q=1 =gt q = 1 - 1/4 = 3/4 = 0.75`

Hence, the probability not to get a correct answer is of `3/4=0.75` .

Evaluating the probability that the student to get 5 correct answers out of 5 such that:

`P(5) = C_5^5*(0.25)^5*(0.75)^(5-5)`

`P(5) = 1*0.0009*1 =gt P(5) = 0.0009`

**Hence, the probability that the student to get 5 answers correct is of 0.0009.**

The student determines the correct answer by choosing a ball from a box that has 4 balls marked with the options A, B, C and D. The probability that a randomly picked ball has the right answer is 0.25.

In a test that has 5 questions the probability that using the system gives all answers right is (0.25)^5 = 9.765*10^-4

**The probability that she gets all five answers right using her system is 9.765*10^-4**