Question

Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is

• A.
• B.
• C.
• D.

Hint:

Drawing without replacement follows Hypergeometric distribution.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We draw 3 oranges from a total of 20 (4 defective + 16 good). The random variable $X$ represents the number of defective oranges. $X$ can take values $\{0, 1, 2, 3\}$.
Step 2: Key Formula or Approach:
Use the hyper-geometric formula (combinations):
$P(X = k) = \frac{\binom{4}{k} \binom{16}{3-k}}{\binom{20}{3}}$.
$\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$.
Step 3: Detailed Explanation:
1. $P(X=0) = \frac{\binom{4}{0} \binom{16}{3}}{1140} = \frac{1 \times 560}{1140} = \frac{28}{57}$.
2. $P(X=1) = \frac{\binom{4}{1} \binom{16}{2}}{1140} = \frac{4 \times 120}{1140} = \frac{480}{1140} = \frac{8}{19}$.
3. $P(X=2) = \frac{\binom{4}{2} \binom{16}{1}}{1140} = \frac{6 \times 16}{1140} = \frac{96}{1140} = \frac{8}{95}$.
4. $P(X=3) = \frac{\binom{4}{3} \binom{16}{0}}{1140} = \frac{4 \times 1}{1140} = \frac{1}{285}$.
Step 4: Final Answer:
The correct distribution is given in Choice (B).