Question

In a multiple choice exam, there are 5 questions and 4 choices for each question (a, b, c, d). Nancy has not studied for the exam at all and decides to randomly guess the answers. What is the probability that:.

Answer 1

Using the binomial distribution, it is found that there is a 0.6836 = 68.36% probability that she guesses at least one right.

For each question, there are only two possible outcomes, either she gets it right or she does not get it right. The probability of getting a question right is independent of any other question, hence the binomial distribution is used to solve this question.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• There are 5 questions, hence n = 5.

• For each question, she guesses one out of 4 options, hence p = 1/4 = 0.25.

The probability that she guesses at least one right is:

$P(X \geq 1) = 1 - P(X = 0)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.25)^{0}.(0.75)^{4} = 0.3164$

Hence:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.3164 = 0.6836$

0.6836 = 68.36% probability that she guesses at least one right.

You can learn more about the binomial distribution at brainly.com/question/24863377