70A bag contains 2 gel pens, 4 ink pens and 6 sketch pens. If two pens were…

Concept: When items are drawn from a bag without replacement, the probability of getting a specific mix across categories equals (the number of ways to choose the required items from each category) divided by (the number of ways to choose the same total count of items from the whole bag) — a single simultaneous draw uses combinations (nCr), not permutations, because order does not matter. When an event needs one item from EACH of two specified categories, the favourable count is the PRODUCT of the two individual category counts, not their sum.

Application:

1. Total pens in the bag = 2 gel + 4 ink + 6 sketch = 12 pens.

2. Total ways to choose any 2 pens out of 12, ignoring order: C(12,2) = (12 × 11) / (2 × 1) = 66.

3. Favourable ways: 1 ink pen chosen from the 4 ink pens gives C(4,1) = 4 ways, and 1 sketch pen chosen from the 6 sketch pens gives C(6,1) = 6 ways; by the multiplication rule, favourable pairs = 4 × 6 = 24.

4. Probability = favourable ÷ total = 24 / 66, which simplifies (divide numerator and denominator by 6) to 4/11.

Cross-check with sequential probability: P(ink first, then sketch) = (4/12) × (6/11) = 24/132, and P(sketch first, then ink) = (6/12) × (4/11) = 24/132. Since either order gives one ink pen and one sketch pen, add the two: 24/132 + 24/132 = 48/132 = 4/11 — the same result as the combinations method.

Answer: The probability that one pen is an ink pen and one is a sketch pen is 4/11.