Question

in a certaib college class 40 percent of students scored an a on the finak exam what is the probabillity pf randomly selecting twi out of six students from the class who scored an a use the formula n^c kp^k q^n-k to solve
A.14 percent
B. 31 percent
C. 4 percent
D. 2 percent

Answer 1

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:

1. Calculate the binomial coefficient \( \binom{n}{k} = \binom{6}{2} \):
\[
\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15
\]

2. Now substitute into the formula:
\[
P(X = 2) = \binom{6}{2} (0.40)^2 (0.60)^{6-2}
\]
\[
P(X = 2) = 15 \cdot (0.40)^2 \cdot (0.60)^4
\]

3. Calculate \( (0.40)^2 = 0.16 \) and \( (0.60)^4 = (0.60 \times 0.60 \times 0.60 \times 0.60) = 0.1296 \):
\[
P(X = 2) = 15 \cdot 0.16 \cdot 0.1296
\]

4. Now perform the multiplication:
\[
15 \cdot 0.16 = 2.4
\]
\[
2.4 \cdot 0.1296 = 0.310656
\]

5. Finally, convert \( 0.310656 \) to a percentage:
\[
0.310656 \times 100 \approx 31.07\%
\]

The nearest answer choice is:

**B. 31 percent**.