Question: In a study of hermit crab behaviour at Point Lookout, North Stradbroke Island, a random sample of 3 types of shells was collected. Each shell was then recorded as being either occupied by a hermit crab or empty. Shells with their original inhabitants were not sampled. The results are shown in the table below:
Shell Type Occupied Empty
Austrocochlea 43 34
Bembicium 9 38
Cirithid 229 47
If we pick a shell at random from this sample and find that it is occupied, the probability that the shell is from a Bembicium is
1. **State the problem:** We want to find the probability that a randomly chosen occupied shell is from the Bembicium type.
2. **Identify the data:**
- Austrocochlea occupied = $43$
- Bembicium occupied = $9$
- Cirithid occupied = $229$
3. **Total occupied shells:**
$$43 + 9 + 229 = 281$$
4. **Formula for conditional probability:**
The probability that the shell is from Bembicium given it is occupied is
$$P(\text{Bembicium} | \text{Occupied}) = \frac{\text{Number of occupied Bembicium shells}}{\text{Total occupied shells}}$$
5. **Calculate the probability:**
$$P = \frac{9}{281}$$
6. **Simplify fraction if possible:**
$9$ and $281$ have no common factors other than $1$, so the fraction is already in simplest form.
7. **Convert to decimal (optional):**
$$P \approx 0.0320$$
**Final answer:** The probability that a randomly chosen occupied shell is from Bembicium is $$\frac{9}{281} \approx 0.0320$$.