Question

There are 20 boys and 5 girls in a class. Three students are selected at random. If $E$ is an event of selecting one boy and two girls, then $P(E)$ is equal to

• A. $\frac{1}{23}$
• B. $\frac{4}{23}$
• C. $\frac{5}{23}$
• D. $\frac{3}{23}$
• E. $\frac{2}{23}$

Hint:

Always simplify your probability fractions at the end. Here, canceling the zeros makes the final fraction $\frac{2}{23}$ clear immediately.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. Here, we use combinations to count the ways.
Key Formula or Approach:
Total ways to pick $r$ from $n$: $\binom{n}{r}$.

Step 2: Detailed Explanation:
Total number of students = $20 + 5 = 25$.
Total number of ways to select 3 students from 25:
\[ n(S) = \binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 2300 \]
Number of ways to select 1 boy from 20 and 2 girls from 5:
\[ n(E) = \binom{20}{1} \times \binom{5}{2} \]
\[ n(E) = 20 \times \frac{5 \times 4}{2 \times 1} = 20 \times 10 = 200 \]
Calculating probability:
\[ P(E) = \frac{n(E)}{n(S)} = \frac{200}{2300} = \frac{2}{23} \]

Step 3: Final Answer:
The probability $P(E)$ is $2/23$.