Question

A test consists of 18 multiple choice questions, each with 3 options. A student who didn’t study randomly guesses on all the questions. What is the probability that she will get a passing score (i.e. 13 out the 20 correct).

Answer 1

Using the binomial distribution, it is found that there is a 0.0008 = 0.08% probability that she will get a passing score.

For each question, there are only two possible outcomes, either she answers it correctly, or she does not. The probability of answering a question correctly is independent of any other question, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

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:

• 18 questions, thus $n = 18$
• Guess one of three correct options, thus $p = \frac{1}{3} = 0.3333$.

A passing score is at least 13 correct, thus, the probability is:

$P(X \geq 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)$

Then

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

$P(X = 13) = C_{18,13}.(0.3333)^{13}.(0.6667)^{5} = 0.0007$

$P(X = 14) = C_{18,14}.(0.3333)^{14}.(0.6667)^{4} = 0.0001$

The others, until P(X = 18), will be approximately 0, thus:

$P(X \geq 13) = P(X = 13) + P(X = 14) = 0.0007 + 0.0001 = 0.0008$

0.0008 = 0.08% probability that she will get a passing score.

A similar problem is given at brainly.com/question/24863377