A landscaping company has a total inventory of ornamental plants consisting only of shrubs, grasses, and trees. If one plant...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{P(shrub)} = \frac{1}{3}\)
  • \(\mathrm{P(grass)} = \frac{1}{2}\)
  • Number of trees = 70
  • Find: Total number of plants

2. INFER the approach

• Since shrubs, grasses, and trees are the only plant types, their probabilities must sum to 1
• Strategy: Find P(tree) first, then use the relationship between probability and count

3. SIMPLIFY to find P(tree)

• Set up the equation: \(\mathrm{P(shrub)} + \mathrm{P(grass)} + \mathrm{P(tree)} = 1\)
• Substitute: \(\frac{1}{3} + \frac{1}{2} + \mathrm{P(tree)} = 1\)
• Find common denominator: \(\frac{2}{6} + \frac{3}{6} + \mathrm{P(tree)} = 1\)
• Combine: \(\frac{5}{6} + \mathrm{P(tree)} = 1\)
• Solve: \(\mathrm{P(tree)} = 1 - \frac{5}{6} = \frac{1}{6}\)

4. INFER the connection between probability and count

• Use the relationship: \(\mathrm{Number\ of\ trees} = \mathrm{Total\ plants} \times \mathrm{P(tree)}\)
• Substitute known values: \(70 = \mathrm{Total} \times \frac{1}{6}\)

5. SIMPLIFY to find total plants

• Solve for total: \(\mathrm{Total} = 70 \times 6 = 420\)

Answer: D. 420

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that all probabilities must sum to 1, so they can't find P(tree). Instead, they might try to directly calculate the total using only the given probabilities, leading to confusion about how to proceed. This causes them to get stuck and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when adding fractions \(\frac{1}{3} + \frac{1}{2}\), incorrectly getting results like \(\frac{2}{5}\) instead of \(\frac{5}{6}\). This gives them \(\mathrm{P(tree)} = -\frac{1}{5}\), which they might reject, or \(\mathrm{P(tree)} = \frac{3}{5}\), leading to a total of \(70 \div \frac{3}{5} = 70 \times \frac{5}{3} \approx 117\). This doesn't match any answer choice exactly, leading to confusion and guessing.

The Bottom Line:

This problem tests whether students understand that probability distributions must be complete (sum to 1) and can connect abstract probability values to concrete counts. The key insight is working backwards from probability to find the total.