Questions: Given that 27% of college students have repeated a course to improve their grade, in a sample of size n=24, use the Binomial Probability formula to find (a) the probability that exactly 8 students have repeated a course. (b) the probability that fewer than 2 students have repeated a course Round answers to the nearest thousandth (3 decimal places). (a) The probability that exactly 8 students have repeated a course is: (b) The probability that fewer than 2 students have repeated a course is

Solution Steps

To solve this problem, we will use the Binomial Probability formula. The binomial probability formula is given by:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where:

• \( n \) is the number of trials (in this case, 24),
• \( k \) is the number of successes (in this case, 8 for part (a) and 0 or 1 for part (b)),
• \( p \) is the probability of success on a single trial (27% or 0.27).

For part (a), we will calculate the probability that exactly 8 students have repeated a course. For part (b), we will calculate the probability that fewer than 2 students have repeated a course, which means we need to find the sum of the probabilities for 0 and 1 student repeating a course.

The binomial probability formula is given by:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

where:

• \( n \) is the number of trials,
• \( k \) is the number of successes,
• \( p \) is the probability of success on a single trial.

Given:

• \( n = 24 \)
• \( p = 0.27 \)
• \( k = 8 \)

We calculate:

\[ P(X = 8) = \binom{24}{8} (0.27)^8 (0.73)^{16} \]

Using the binomial coefficient and the given values, we find:

\[ P(X = 8) \approx 0.1351 \]

We need to find the sum of the probabilities for \( k = 0 \) and \( k = 1 \):

\[ P(X < 2) = P(X = 0) + P(X = 1) \]

For \( k = 0 \):

\[ P(X = 0) = \binom{24}{0} (0.27)^0 (0.73)^{24} \approx 0.0052 \]

For \( k = 1 \):

\[ P(X = 1) = \binom{24}{1} (0.27)^1 (0.73)^{23} \approx 0.0000 \]

Summing these probabilities:

\[ P(X < 2) \approx 0.0052 \]

Final Answer