A bag contains 4 white balls, 5 red balls and 6 blue balls. Three balls are drawn at random without replacement. What is the probability that all the three balls drawn are red?

Introduction / Context:
This problem is a classic example of probability without replacement and uses combinations. It asks for the probability that all three chosen balls are red when drawing from a bag containing balls of three different colours. The focus is on counting combinations correctly and understanding that the order of drawing does not matter in this context.

Given Data / Assumptions:

• White balls = 4.
• Red balls = 5.
• Blue balls = 6.
• Total balls = 4 + 5 + 6 = 15.
• Three balls are drawn at random without replacement.
• Event of interest: all three drawn balls are red.

Concept / Approach:
Since order does not matter, we use combinations. The total number of possible selections of 3 balls is C(15, 3). The number of favourable selections, where every ball is red, is C(5, 3), because we must choose all three from the 5 red balls. The required probability is then given by favourable combinations divided by total combinations.

Step-by-Step Solution:
Total balls = 15, so total combinations = C(15, 3). C(15, 3) = 15 * 14 * 13 / (3 * 2 * 1) = 455. Red balls = 5, so favourable combinations = C(5, 3). C(5, 3) = 5 * 4 * 3 / (3 * 2 * 1) = 10. Required probability = 10 / 455. Simplify 10 / 455 by dividing numerator and denominator by 5: 10 / 455 = 2 / 91.

Verification / Alternative check:
We can also compute stepwise probability without combinations. First draw red: 5/15. Second red: 4/14. Third red: 3/13. Multiply: (5/15) * (4/14) * (3/13) = (5 * 4 * 3) / (15 * 14 * 13) = 60 / 2730 = 2 / 91 after simplification. This matches the combination method answer.

Why Other Options Are Wrong:
1/22 and 3/22 are larger than 2/91 and do not match the correct counting. 2/77 corresponds to 2 * 455 / (77 * 455), which does not arise from any correct favourable count.

Common Pitfalls:
Students sometimes mix permutations and combinations, or forget that the total number of balls is 15, not 5. Another error is computing probabilities additively rather than multiplicatively for successive draws. Always check whether order matters and whether replacement occurs.

Final Answer:
So, the probability that all three drawn balls are red is 2/91.