A box contains 5 defective bulbs and 15 non defective bulbs, for a total of 20 bulbs. Two bulbs are chosen at random without replacement. What is the probability that both bulbs chosen are non defective?

Introduction / Context:
This question tests probability with selections from a finite population containing defective and non defective items. You are required to find the probability that both chosen bulbs are non defective when the choice is made without replacement, which means each draw changes the composition of the box.

Given Data / Assumptions:

• Number of defective bulbs = 5.
• Number of non defective bulbs = 15.
• Total bulbs = 5 + 15 = 20.
• Two bulbs are drawn at random without replacement.
• We want the probability that both bulbs are non defective.

Concept / Approach:
The probability can be computed either stepwise using conditional probabilities or using combinations. With combinations, we compare the number of ways to select 2 non defective bulbs to the number of ways to select any 2 bulbs from the total. The formula is: Probability = (Number of ways to choose 2 non defective bulbs) / (Number of ways to choose any 2 bulbs).

Step-by-Step Solution:
Step 1: Total number of ways to choose any 2 bulbs from 20 = 20C2. 20C2 = 20 * 19 / (2 * 1) = 190. Step 2: Total number of non defective bulbs = 15. Step 3: Number of ways to choose 2 non defective bulbs = 15C2. 15C2 = 15 * 14 / (2 * 1) = 105. Step 4: Probability that both bulbs are non defective = 105 / 190. Step 5: Simplify 105 / 190 by dividing numerator and denominator by 5 to get 21 / 38.

Verification / Alternative check:
You can also use a stepwise approach. The probability that the first bulb is non defective is 15/20 = 3/4. After one non defective bulb is taken, the remaining non defective bulbs are 14 and the total bulbs are 19. The probability that the second bulb is also non defective is 14/19. Therefore, probability both are non defective is: (3/4) * (14/19) = 42 / 76 = 21 / 38 after simplification. This matches the combination method exactly.

Why Other Options Are Wrong:
Option 15/38 and option 19/38 are both smaller than 21/38 and do not result from any correct combination or stepwise multiplication. Option 3/19 is much too small, and option 5/19 does not match either of the systematic methods. Only 21/38 aligns with both the combination based and the conditional probability based calculations.

Common Pitfalls:
A common mistake is to treat the draws as independent with replacement and mistakenly use (15/20)^2 with no adjustment, which actually does equal 9/16 and is different from the correct result. Another error is to use 15C2 in the numerator but incorrectly put 20^2 or 20C1 * 19C1 in the denominator without proper reasoning. Always confirm whether the selection is with or without replacement and then be consistent in your use of combinations or sequential probabilities, not mixing concepts from different models.

Final Answer:
The probability that both bulbs drawn are non defective is 21/38.