Question

In Exercises 15–20, assume that random guesses are made for eight multiple choice questions on an SAT test, so that there are n = 8 trials, each with probability of success (correct) given by p = 0.20. Find the indicated probability for the number of correct answers.

Find the probability that the number x of correct answers is at least 4.

Short answer

The probability of getting at least four correct answers is equal to 0.056282.

Step-by-step solution

Given information

A set of eight multiple-choice questions are answered in the SAT. The probability of a correct answer is given to be equal to 0.20.

Calculate the required probability

Let X denote the number of correct answers.

Let success be defined as getting a correct answer.

Thus, the number of trials (n) is given to be equal to eight.

The probability of success (getting a correct answer) is p= 0.20.

The probability of failure (getting a wrong answer) is calculated below:

$$\begin{gathered} q=1-p \\ =1-0.20 \\ =0.80 \end{gathered}$$

The number of successes required in eight trials should be at least four.

The binomial probability formula is as follows:

$P\left(X=x\right)={}^n{C}_x{\left(p\right)}^x{\left(q\right)}^{n-x}$

By using the binomial probability formula, the probability of getting seven correct answers is computed below:

$$\begin{gathered} P\left(X⩾4\right)=1-P\left(X<4\right) \\ =1-\left(P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)\right) \\ =1-\left({}^8{C}_0{\left(0.20\right)}^0{\left(0.80\right)}^8+{}^8{C}_1{\left(0.20\right)}^1{\left(0.80\right)}^7+{}^8{C}_2{\left(0.20\right)}^2{\left(0.80\right)}^6+{}^8{C}_3{\left(0.20\right)}^3{\left(0.80\right)}^5\right) \\ =1-0.943718 \end{gathered}$$

$=0.056282$

Therefore, the probability of getting at least four correct answers is equal to 0.056282.