Question

A group consists of 9 students from 3 disciplines. Out of the 9 students, 4 are from Statistics, 3 are from Economics and 2 are from Chemistry. Three students are selected randomly. What is the probability that, out of the selected students 2 are from Statistics?

• A. $\frac{3}{14}$
• B. $\frac{2}{14}$
• C. $\frac{5}{14}$
• D. $\frac{1}{14}$

Hint:

Group all non-target categories together (e.g., Economics and Chemistry $\rightarrow$ Non-Statistics) to simplify the calculation into a standard binary-choice combination.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem represents a classical hypergeometric selection model, where items are drawn without replacement from a finite population with distinct subgroups.
Key Formula or Approach:
The probability of selecting exactly $k$ items of a specific category out of a total selection of $n$ from a population of size $N$ containing $M$ items of that category is:
\[ P(X = k) = \frac{\binom{M}{k} \binom{N - M}{n - k}}{\binom{N}{n}} \]

Step 2: Detailed Explanation:
Let us identify the parameters of the problem:
Total students, $N = 9$
Target category (Statistics) students, $M = 4$
Non-Statistics students (Economics + Chemistry), $N - M = 3 + 2 = 5$
Sample size drawn, $n = 3$
We want to find the probability of selecting exactly $k = 2$ Statistics students.
Calculate the combinations:
1. Ways to select 2 Statistics students from 4:
\[ \binom{4}{2} = \frac{4 \times 3}{2} = 6 \]
2. Ways to select remaining $3 - 2 = 1$ student from the 5 non-statistics students:
\[ \binom{5}{1} = 5 \]
3. Total ways to select 3 students out of 9:
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84 \]
Substitute these values into the probability formula:
\[ P(X = 2) = \frac{\binom{4}{2} \binom{5}{1}}{\binom{9}{3}} = \frac{6 \times 5}{84} = \frac{30}{84} \]
Simplify the fraction by dividing the numerator and denominator by 6:
\[ P(X = 2) = \frac{5}{14} \]
Therefore, the probability is $\frac{5}{14}$.

Step 3: Final Answer
The correct option is (C).