Question

In a certain college class, 40 percent of students scored an A on the final exam. What is the probability of randomly selecting two out of six students from the class who scored an A? Use the formula nCkp" a-k to solve.

Answer 1

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:

\[
nCk = \frac{n!}{k!(n-k)!}
\]

1. Calculate \( nCk \):

\[
nCk = \frac{6!}{2!(6-2)!} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5}{2 \times 1} = 15
\]

2. Calculate \( p^k \) and \( q^{n-k} \):

\[
p^k = (0.4)^2 = 0.16
\]
\[
q^{(n-k)} = (0.6)^{(6-2)} = (0.6)^4 = 0.1296
\]

3. 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)