Question

A box contains 5 defective and 15 non-defective bulbs. Two bulbs are drawn at random without replacement. What is the probability that at least one bulb is defective?

• A. \(\frac{17}{38}\)
• B. \(\frac{7}{20}\)
• C. \(\frac{3}{5}\)
• D. \(\frac{17}{20}\)

Hint:

When asked "at least one" in probability, use the complement rule: \[ P(\text{at least one}) = 1 - P(\text{none}). \]

Answer 1

The Correct Option is A

Step 1: Total bulbs = \(5 + 15 = 20.\) 
Step 2: Calculate the probability that at least one bulb is defective. 
This is easier to find using the complement: \[ P(\text{at least one defective}) = 1 - P(\text{no defective}). \] 
Step 3: Calculate \(P(\text{no defective})\), i.e., both bulbs are non-defective. 
Number of non-defective bulbs = 15. 
Number of ways to pick 2 non-defective bulbs without replacement: \[ P(\text{both non-defective}) = \frac{15}{20} \times \frac{14}{19} = \frac{15 \times 14}{20 \times 19} = \frac{210}{380} = \frac{21}{38}. \] 
Step 4: Therefore, \[ P(\text{at least one defective}) = 1 - \frac{21}{38} = \frac{38 - 21}{38} = \frac{17}{38}. \] 
Step 5: Simplify \(\frac{17}{38}\). Since \(17\) and \(38\) have no common factors other than 1, \(\frac{17}{38}\) is already in simplest form. 

Answer 2

Alternative method: Calculate directly the probability of at least one defective bulb: \[ P(\text{exactly one defective}) = \frac{5}{20} \times \frac{15}{19} + \frac{15}{20} \times \frac{5}{19} = \frac{5 \times 15}{20 \times 19} + \frac{15 \times 5}{20 \times 19} = \frac{75}{380} + \frac{75}{380} = \frac{150}{380} = \frac{15}{38}. \] \[ P(\text{exactly two defective}) = \frac{5}{20} \times \frac{4}{19} = \frac{20}{380} = \frac{1}{19}. \] Sum these: \[ P(\text{at least one defective}) = \frac{15}{38} + \frac{1}{19} = \frac{15}{38} + \frac{2}{38} = \frac{17}{38} \approx 0.447, \] which is consistent. 
Conclusion: The correct probability is \(\frac{17}{38}\).