Question

On a multiple-choice exam with three possible answers for each of the five questions, what is the probability that a student would get four or more correct answers just by guessing?

Answer 1

Answer:

The required probability is $\frac{11}{243}$.

Step-by-step explanation:

Consider the provided information.

On a multiple-choice exam with three possible answers for each of the five questions.

That means the Probability of guess being correct is = $p=\frac{1}{3}$

The Probability of guess being incorrect is = $q=\frac{2}{3}$

We need to find the probability that a student would get four or more correct answers just by guessing.

Let X represents the probability of being correct

$P(X\geq4)=P(X=4)+P(X=5)$

Use the binomial distribution: $^nC_r=\binom{n}{r}p^rq^{n-r}$

$P(X\geq4)=\binom{5}{4}(\frac{1}{3})^4(\frac{2}{3})^{1}+\binom{5}{5}(\frac{1}{3})^5(\frac{2}{3})^{0}$

$P(X\geq4)=5\times(\frac{1}{3})^4(\frac{2}{3})^{1}+(\frac{1}{3})^5\\\\P(X\geq4)=\frac{10}{243}+\frac{1}{243}\\\\P(X\geq4)=\frac{11}{243}$

Hence, the required probability is $\frac{11}{243}$.