To solve this problem, we can model the situation using a binomial distribution. We define success as a student graduating high school (which has a probability of \( p = 1 - 0.103 = 0.897 \)). The number of trials is \( n = 10 \), as we are choosing 10 students.
We want to find the probability that at least 6 students graduate. This can be expressed as:
\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]
where \( X \) is the number of students graduating and follows a binomial distribution \( X \sim B(n = 10, p = 0.897) \).
The probability mass function of a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient.
Now, we can calculate each of these probabilities:
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For \( k = 6 \): \[ P(X = 6) = \binom{10}{6} (0.897)^6 (0.103)^4 \] \[ = 210 \cdot (0.897)^6 \cdot (0.103)^4 \approx 210 \cdot 0.536 \cdot 0.000115 \approx 0.012505 \]
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For \( k = 7 \): \[ P(X = 7) = \binom{10}{7} (0.897)^7 (0.103)^3 \] \[ = 120 \cdot (0.897)^7 \cdot (0.103)^3 \approx 120 \cdot 0.479 \cdot 0.001092 \approx 0.062887 \]
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For \( k = 8 \): \[ P(X = 8) = \binom{10}{8} (0.897)^8 (0.103)^2 \] \[ = 45 \cdot (0.897)^8 \cdot (0.103)^2 \approx 45 \cdot 0.428 \cdot 0.010609 \approx 0.182486 \]
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For \( k = 9 \): \[ P(X = 9) = \binom{10}{9} (0.897)^9 (0.103)^1 \] \[ = 10 \cdot (0.897)^9 \cdot (0.103)^1 \approx 10 \cdot 0.383 \cdot 0.103 \approx 0.394065 \]
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For \( k = 10 \): \[ P(X = 10) = \binom{10}{10} (0.897)^{10} (0.103)^0 \] \[ = 1 \cdot (0.897)^{10} \approx 0.343 \]
Now, we sum these probabilities:
\[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \]
Calculating the total:
\[ P(X \geq 6) \approx 0.012505 + 0.062887 + 0.182486 + 0.394065 + 0.343963 \approx 0.995906 \]
Rounding to the nearest thousandth:
\[ P(X \geq 6) \approx 0.996 \]
Thus, the probability that at least 6 of the 10 randomly chosen students graduate is approximately 0.996.
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