Question

A packet contains 10 distinguishable firecrackers out of which 4 are defective. If three firecrackers are drawn at random (without replacement) from the packet, then the probability that all three firecrackers are defective equals

• A. $\dfrac{1}{10}$
• B. $\dfrac{1}{20}$
• C. $\dfrac{1}{30}$
• D. $\dfrac{1}{40}$

Hint:

In problems involving "without replacement", use combinations $\binom{n}{r}$ for both total and favorable outcomes.

Answer 1

The Correct Option is C

Step 1: Total and favorable combinations.
Total ways to choose 3 out of 10: $\binom{10}{3} = 120$. Favorable ways to choose 3 defective out of 4: $\binom{4}{3} = 4$.

Step 2: Compute probability.
\[ P(\text{all defective}) = \frac{\binom{4}{3}}{\binom{10}{3}} = \frac{4}{120} = \frac{1}{30}. \]

Step 3: Conclusion.
\[ \boxed{P = \frac{1}{30}}. \]