To find the probability of randomly selecting two out of six students who scored an A on the final exam, we can use the binomial probability formula. The problem states that 40% of the students scored an A. Therefore, we can set this up as a binomial distribution problem where:
- \( n = 6 \): the number of trials (students),
- \( k = 2 \): the number of successes (students who scored an A),
- \( p = 0.4 \): the probability of success (scoring an A),
- \( q = 1 - p = 0.6 \): the probability of failure (not scoring an A).
The binomial probability formula is given by:
\[ P(X = k) = nCk \cdot p^k \cdot q^{(n-k)} \]
Where \( nCk \) is the binomial coefficient, computed as:
\[ nCk = \frac{n!}{k!(n-k)!} \]
- Calculate \( nCk \):
\[ nCk = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5}{2 \times 1} = 15 \]
- Calculate \( p^k \) and \( q^{n-k} \):
\[ p^k = (0.4)^2 = 0.16 \] \[ q^{(n-k)} = (0.6)^{(6-2)} = (0.6)^4 = 0.1296 \]
- Combine all parts into the probability formula:
\[ P(X = 2) = 15 \cdot 0.16 \cdot 0.1296 \]
Calculating this step-by-step:
\[ 15 \cdot 0.16 = 2.4, \] \[ 2.4 \cdot 0.1296 = 0.310656 \]
Thus, the probability of randomly selecting two out of six students from the class who scored an A is:
\[ \boxed{0.3107} \] (rounding to four decimal places)
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