To solve this problem, we can use the binomial probability formula, which is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Where:
- \( n \) is the total number of trials (students selected)
- \( k \) is the number of successful outcomes (students scoring an A)
- \( p \) is the probability of success (probability a student scores an A)
- \( q \) is the probability of failure (probability a student does not score an A)
- \( \binom{n}{k} \) is the binomial coefficient which represents the number of ways to choose \( k \) successes in \( n \) trials.
In your case, we are selecting \( n = 6 \) students, and we want to find the probability of \( k = 2 \) students scoring an A.
Given:
- \( p = 0.40 \) (the probability of scoring an A)
- \( q = 1 - p = 0.60 \)
Now, we'll calculate the probability that exactly 2 out of 6 students scored an A:
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Calculate the binomial coefficient \( \binom{n}{k} = \binom{6}{2} \): \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \]
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Now substitute into the formula: \[ P(X = 2) = \binom{6}{2} (0.40)^2 (0.60)^{6-2} \] \[ P(X = 2) = 15 \cdot (0.40)^2 \cdot (0.60)^4 \]
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Calculate \( (0.40)^2 = 0.16 \) and \( (0.60)^4 = (0.60 \times 0.60 \times 0.60 \times 0.60) = 0.1296 \): \[ P(X = 2) = 15 \cdot 0.16 \cdot 0.1296 \]
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Now perform the multiplication: \[ 15 \cdot 0.16 = 2.4 \] \[ 2.4 \cdot 0.1296 = 0.310656 \]
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Finally, convert \( 0.310656 \) to a percentage: \[ 0.310656 \times 100 \approx 31.07% \]
The nearest answer choice is:
B. 31 percent.
1 answer