To find the probability that exactly 17 out of 29 high school graduates enroll in college, we can use the binomial probability formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where:
- \( n \) is the number of trials (which is 29),
- \( k \) is the number of successes (which is 17 in this case),
- \( p \) is the probability of success on a single trial (which is 0.65 since 65% of students enroll),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).
Using the values provided:
- \( n = 29 \)
- \( k = 17 \)
- \( p = 0.65 \)
First, calculate \( \binom{29}{17} \):
\[ \binom{29}{17} = \frac{29!}{17!(29-17)!} = \frac{29!}{17! \cdot 12!} \]
Calculating \( 29! \), \( 17! \), and \( 12! \):
- \( 29! = 29 \times 28 \times \ldots \times 1 \)
- \( 17! = 17 \times 16 \times \ldots \times 1 \)
- \( 12! = 12 \times 11 \times \ldots \times 1 \)
However, you can simplify the calculation without calculating factorials directly:
\[ \binom{29}{17} = \frac{29 \times 28 \times ... \times 13}{17 \times 16 \times ... \times 1} \]
Upon calculation, we find:
\[ \binom{29}{17} = 17383860 \]
Now substitute the values into the binomial formula:
\[ P(X = 17) = \binom{29}{17} (0.65)^{17} (0.35)^{12} \]
Calculating \( (0.65)^{17} \):
\[ (0.65)^{17} \approx 0.020429 \]
Calculating \( (0.35)^{12} \):
\[ (0.35)^{12} \approx 0.000024884 \]
Now putting these values into the formula:
\[ P(X = 17) \approx 17383860 \times 0.020429 \times 0.000024884 \approx 0.857581 \]
Finally, rounding the answer to four decimal places gives us:
\[ P(X = 17) \approx 0.3185 \]
So, the probability that exactly 17 out of the 29 high school graduates enroll in college is approximately 0.3185.
1 answer