A bag contains 7 green balls and 5 black balls. Three balls are drawn one after another without replacement. What is the probability that all the three balls drawn from the bag are green?

Introduction / Context:
This question tests your understanding of basic probability for selections made without replacement. The key idea is to calculate the probability that all three balls taken from a finite collection are green when each draw changes the composition of the bag. Problems like this are very common in quantitative aptitude exams and help you practise sequential probability and careful fraction handling.

Given Data / Assumptions:

• Total green balls in the bag = 7
• Total black balls in the bag = 5
• Total balls in the bag initially = 12
• Three balls are drawn one after another
• Draws are without replacement
• We want the probability that all three drawn balls are green

Concept / Approach:
The general approach is to treat each draw as a conditional probability step. For draws without replacement, the denominator (total balls) and the numerator (favorable balls) change after every draw. We multiply the probabilities for each draw because all three events must occur together: green on the first draw, green on the second draw, and green on the third draw. This is an example of using the multiplication rule for dependent events.

Step-by-Step Solution:
Step 1: Initially there are 12 balls in total and 7 of them are green. Step 2: Probability that the first ball is green = 7/12. Step 3: After drawing one green ball, remaining green balls = 6 and total balls = 11. Step 4: Probability that the second ball is green, given that the first was green = 6/11. Step 5: After drawing two green balls, remaining green balls = 5 and total balls = 10. Step 6: Probability that the third ball is green, given that first two were green = 5/10. Step 7: Multiply all three probabilities to get the overall probability. Overall probability = (7/12) * (6/11) * (5/10). Step 8: Simplify: (7 * 6 * 5) / (12 * 11 * 10) = 210 / 1320. Step 9: Divide numerator and denominator by 30 to simplify: 210/1320 = 7/44.

Verification / Alternative check:
Another way is to count combinations. The number of ways to choose any 3 balls from 12 is 12C3. The number of ways to choose 3 green balls from the 7 green balls is 7C3. The probability is 7C3 / 12C3. We have 7C3 = 35 and 12C3 = 220, so probability = 35/220 = 7/44, which matches the stepwise multiplication result. This confirms that our calculation is consistent and correct.

Why Other Options Are Wrong:
Option 21/88 is obtained if the denominators or simplification are handled incorrectly, so it is too small. Option 7/24 is larger than the true probability and effectively assumes more favorable outcomes than actually exist. Option 5/22 does not arise from any correct systematic calculation with either the stepwise or combination method. Option 3/11 is also an overestimate and does not correspond to the true reduced fraction of 7/44.

Common Pitfalls:
Many learners mistakenly treat draws as independent and keep the denominator fixed at 12 for all three draws, which is wrong for selection without replacement. Another frequent error is to use combinations for the numerator and simple powers like 12^3 for the denominator, which would correspond to ordered draws with replacement. Students also sometimes forget to simplify fractions correctly or cancel common factors in the wrong way. Always track how many favorable balls and how many total balls remain after each draw, and be disciplined in your simplification steps.

Final Answer:
The probability that all three balls drawn are green is 7/44.