To find the probability that no more than two out of ten randomly selected American high school students drop out, we can 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 (students, in this case).
- \( k \) is the number of successes (students that drop out).
- \( p \) is the probability of success (the dropout rate).
- \( \binom{n}{k} \) is the binomial coefficient.
Given:
- \( n = 10 \)
- \( p = 0.103 \) (the dropout probability)
- \( 1 - p = 0.897 \) (the probability of not dropping out)
We want to find the probability that no more than two students drop out, which is \( P(X \leq 2) \). This can be calculated as:
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
Now, we will compute each of these probabilities:
- Calculate \( P(X = 0) \):
\[ P(X = 0) = \binom{10}{0} (0.103)^0 (0.897)^{10} = 1 \cdot 1 \cdot (0.897)^{10} \]
Calculating \( (0.897)^{10} \):
\[ (0.897)^{10} \approx 0.3487 \]
So:
\[ P(X = 0) \approx 0.3487 \]
- Calculate \( P(X = 1) \):
\[ P(X = 1) = \binom{10}{1} (0.103)^1 (0.897)^{9} = 10 \cdot 0.103 \cdot (0.897)^{9} \]
Calculating \( (0.897)^{9} \):
\[ (0.897)^{9} \approx 0.3881 \]
So:
\[ P(X = 1) \approx 10 \cdot 0.103 \cdot 0.3881 \approx 0.3993 \]
- Calculate \( P(X = 2) \):
\[ P(X = 2) = \binom{10}{2} (0.103)^2 (0.897)^{8} = 45 \cdot (0.103)^2 \cdot (0.897)^{8} \]
Calculating \( (0.897)^{8} \):
\[ (0.897)^{8} \approx 0.4328 \]
Now calculate \( (0.103)^2 \):
\[ (0.103)^2 \approx 0.010609 \]
So:
\[ P(X = 2) \approx 45 \cdot 0.010609 \cdot 0.4328 \approx 0.2069 \]
- Now add these probabilities together:
\[ P(X \leq 2) \approx P(X = 0) + P(X = 1) + P(X = 2) \approx 0.3487 + 0.3993 + 0.2069 \approx 0.9549 \]
Rounding the final result to the nearest thousandth, the probability that no more than two students drop out is approximately:
\[ \boxed{0.955} \]
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