A driver’s knowledge test contains 30 multiple-choice questions, each with 4 options - HSC - SSCE Mathematics Extension 1 - Question 7 - 2024 - Paper 1

The applicant has 4 options for each question, hence the probability of getting a question right by random guessing is ( p = \frac{1}{4} ). Conversely, the probability of getting a question wrong is ( q = \frac{3}{4} ).

To find the probability of passing, we need to consider two scenarios: getting exactly 4 correct answers and getting all 5 correct.

1. Getting exactly 4 correct: This can happen in ( \binom{5}{4} ) ways:

   ( P(X = 4) = \binom{5}{4} p^4 q^1 = 5 \left(\frac{1}{4}\right)^4 \left(\frac{3}{4}\right)^1 = 5 \cdot \frac{1}{256} \cdot \frac{3}{4} = \frac{15}{256} )

2. Getting all 5 correct:

   ( P(X = 5) = \binom{5}{5} p^5 = 1 \cdot \left(\frac{1}{4}\right)^5 = \frac{1}{1024} )

Now, the total probability of passing is the sum of these two probabilities:

( P(pass) = \frac{15}{256} + \frac{1}{1024} )

To combine these fractions, convert ( \frac{15}{256} ) to a denominator of 1024:

( P(pass) = \frac{60}{1024} + \frac{1}{1024} = \frac{61}{1024} )

The probability of passing the test after randomly guessing the last 5 questions is therefore ( \frac{61}{1024} ), which approximates closely to the original options with the closest correct choice being C.