On a multiple choice test with 20 questions, each question has 5 possible answers, one of which is correct. If a student does not study and guesses on all questions, what is the probability that they get exactly 12 correct? (Hint: there are 2 outcomes, either they get it correct or they get it incorrect)

1 answer

To find the probability that a student gets exactly 12 questions correct on a 20-question multiple choice test where each question has 5 answers, we can model this situation using the binomial probability formula.

In this scenario:

  • The number of trials (questions) \( n = 20 \).
  • The number of successes (correct answers) \( k = 12 \).
  • The probability of success on a single trial (getting a question right) \( p = \frac{1}{5} \).
  • The probability of failure (getting a question wrong) \( q = 1 - p = \frac{4}{5} \).

The binomial probability formula is given by:

\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

Where:

  • \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

Now substituting in our values:

  1. Calculate the binomial coefficient \( \binom{20}{12} \):

\[ \binom{20}{12} = \frac{20!}{12!(20-12)!} = \frac{20!}{12!8!} \]

Calculating \( \binom{20}{12} \):

This can also be expressed as \( \binom{20}{12} = \binom{20}{8} \) due to the property of binomial coefficients.

Using \( \binom{20}{8} \):

\[ \binom{20}{8} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 125970 \]

  1. Next, calculate \( p^{12} \) and \( q^{8} \):

\[ p^{12} = \left(\frac{1}{5}\right)^{12} = \frac{1}{5^{12}} = \frac{1}{244140625} \]

\[ q^{8} = \left(\frac{4}{5}\right)^{8} = \frac{4^8}{5^8} = \frac{65536}{390625} \]

Now, we can substitute these values into the binomial probability formula:

\[ P(X = 12) = \binom{20}{12} \cdot p^{12} \cdot q^{8} = 125970 \cdot \frac{1}{244140625} \cdot \frac{65536}{390625} \]

Calculating the product step-by-step:

\[ P(X = 12) = 125970 \cdot \frac{65536}{244140625 \cdot 390625} \]

Calculating \( 244140625 \times 390625 = 95367431640625 \):

Notice:

\[ P(X = 12) = 125970 \cdot \frac{65536}{95367431640625} \]

Now computing \( 125970 \cdot 65536 = 8251604480 \):

So finally:

\[ P(X = 12) = \frac{8251604480}{95367431640625} \]

This value can be calculated but is more conveniently left as a fraction. Evaluating this:

\[ P(X = 12) \approx 0.00008656 \]

Therefore, the final probability is:

\[ \boxed{0.0000866} \text{ or approximately 0.00866%} \]