Question: A bag has 13 counters in, 8 are blue and the rest are red.
A counter is removed, not replaced and another chosen.
Find the probability that a blue counter is not chosen.
1. **State the problem:** We have a bag with 13 counters: 8 blue and 5 red (since 13 - 8 = 5).
2. **What is asked?** Find the probability that a blue counter is not chosen in two draws without replacement.
3. **Key idea:** "Blue counter is not chosen" means both counters drawn are red.
4. **Calculate the probability:**
- Total counters initially: 13
- Red counters initially: 5
5. **Probability first counter is red:**
$$P(\text{red}_1) = \frac{5}{13}$$
6. **After removing one red counter, counters left:**
- Total counters left: 12
- Red counters left: 4
7. **Probability second counter is red given first was red:**
$$P(\text{red}_2 | \text{red}_1) = \frac{4}{12} = \frac{1}{3}$$
8. **Probability both counters are red:**
$$P(\text{red}_1 \cap \text{red}_2) = P(\text{red}_1) \times P(\text{red}_2 | \text{red}_1) = \frac{5}{13} \times \frac{1}{3} = \frac{5}{39}$$
9. **Therefore, the probability that a blue counter is not chosen is:**
$$\boxed{\frac{5}{39}}$$