Cards numbered from 1, 3, 5, 7, 9, …, 33 are put in a bag. A card is drawn…

Concept: For equally likely outcomes, P(event) = n(favourable outcomes) / n(total outcomes). When the sample space is an arithmetic sequence with first term a, common difference d, and last term l, its size is (l − a)/d + 1.

Application:

1. The bag contains the odd numbers from 1 to 33: an arithmetic progression with first term 1, common difference 2, last term 33, so n(S) = (33 − 1)/2 + 1 = 17.

2. Since every card is odd, no card is divisible by 2, so the event “not divisible by 3 or divisible by 2” reduces to just “not divisible by 3”, because the “divisible by 2” branch contributes zero extra cards.

3. The multiples of 3 among these 17 odd numbers are 3, 9, 15, 21, 27, 33 — exactly 6 numbers.

4. So the numbers not divisible by 3 = 17 − 6 = 11.

5. Probability = 11/17.

Cross-check: Listing the numbers not divisible by 3 directly — {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31} — gives 11 numbers, confirming the count independently.

∴ The required probability is 11/17.