A room contains 3 brown chairs, 5 black chairs and 4 white chairs. Two chairs are picked at random without replacement and placed in the lawn. What is the probability that none of the chairs picked is white?

Introduction / Context:
This question tests probability with selections from a finite set of objects of different types. We have chairs of three different colours and we pick two chairs at random without replacement. The requirement is that none of the chairs picked is white, so both selected chairs must be non white. This can be solved conveniently using combinations.

Given Data / Assumptions:

Brown chairs = 3.

Black chairs = 5.

White chairs = 4.

Total chairs in the room = 3 + 5 + 4 = 12.

Two chairs are chosen without replacement.

We want the probability that neither selected chair is white (both chairs are from brown or black).

Concept / Approach:
We treat the selection as a two chair combination from the total chairs. The total number of ways to choose any two chairs from 12 is 12C2. The favourable selections are those in which both chairs come from the non white group. The non white chairs are the brown and black ones. We count the ways of choosing 2 chairs from this non white group and divide by the total number of combinations.

Step-by-Step Solution:
Step 1: Total number of chairs = 3 brown + 5 black + 4 white = 12 chairs.Step 2: Total number of ways to choose 2 chairs from 12 = 12C2.Step 3: Compute 12C2 = (12 * 11) / (2 * 1) = 66.Step 4: Non white chairs are the brown and black chairs. Number of non white chairs = 3 + 5 = 8.Step 5: Number of ways to choose 2 chairs from the 8 non white chairs = 8C2.Step 6: Compute 8C2 = (8 * 7) / (2 * 1) = 28.Step 7: Probability that none of the chairs picked is white = favourable / total = 28 / 66.Step 8: Simplify 28 / 66 by dividing numerator and denominator by 2 to get 14 / 33.

Verification / Alternative check:
We can verify using sequential probabilities. For the first chair, probability that it is non white is 8/12 = 2/3. Given that the first chair was non white, there are now 7 non white chairs and 11 chairs in total, so probability that the second chair is also non white is 7/11. The combined probability is (2/3) * (7/11) = 14 / 33, which matches the combination method result.

Why Other Options Are Wrong:
The fraction 14/55 uses the wrong total number of two chair combinations and would correspond to a different total number of chairs. The value 12/55 similarly miscalculates both favourable and total outcomes. The fraction 13/33 and 11/33 are larger or smaller than the correct value and do not match either the combinational or sequential probability calculations.

Common Pitfalls:
Students sometimes mistakenly treat the white chairs as favourable instead of unfavourable or miscount the non white chairs. Others try to compute probabilities by subtracting from 1 but then incorrectly count the event that at least one chair is white. Directly counting the all non white pairs is simpler and less error prone here.

Final Answer:
The probability that none of the chairs picked is white is 14/33.