Question 7

A driver’s knowledge test contains 30 multiple-choice questions, each with 4 options. An applicant must get at least 29 correct to pass.

If an applicant correctly answers the first 25 questions and randomly guesses the last 5 questions, what is the probability that the applicant will pass the test?

A. 1256\displaystyle \frac1{256}

B. 151024\displaystyle \frac{15}{1024}

C. 164\displaystyle \frac1{64}

D. 21256\displaystyle \frac{21}{256}

Solution

Using this strategy, the applicant must guess at least 4 of the 5 remaining questions correctly to pass the test. If XX is the random variable which represents the number of correct answers they guess correctly from the last 55 questions, then we can model this as a binomial variable. Since there are 55 questions, the parameter n=5n = 5 for this model, and because each question has 44 options, the the probability of answering any individual question correctly is 14\displaystyle \frac14 if we are guessing randomly, so this is our parameter pp.

We have been asked to calculate P(X≥4)P(X \geq 4), and so we can use the formula for the binomial distribution to calculate this probability

P(X=x)=5Cx(14)x(1−14)5−xP(X=x) = {}^5C_x \left(\frac14\right)^x \left(1-\frac14\right)^{5-x}

The probability of guessing all five answers correctly is equal to

P(X=5)=5C5(14)5(1−14)5−5P(X=5)=1⋅(14)5⋅1P(X=5)=(14)5\begin{aligned} P(X=5) = {}^5C_5 \left(\frac14\right)^5 \left(1-\frac14\right)^{5-5} \\ P(X=5) = 1 \cdot \left(\frac14\right)^5 \cdot 1 \\ P(X=5) = \left(\frac14\right)^5 \\ \end{aligned}

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