Two classes at a school are labeled Class A and Class B. Class A has 24 students, of whom 10...

1. TRANSLATE the problem information

• Given information:
  • Class A: 24 students, 10 left-handed
  • Class B: 30 students, 9 left-handed
  • One class chosen randomly, then one student chosen randomly from that class
  • The selected student IS left-handed (this is key!)

• We need to find: Probability the student came from Class B, given they're left-handed

2. INFER the correct approach

• This is asking for \(\mathrm{P(Class\ B\ |\ Left-handed)}\), not \(\mathrm{P(Left-handed\ |\ Class\ B)}\)
• This classic "reverse probability" requires Bayes' theorem
• We'll need the law of total probability to find \(\mathrm{P(Left-handed)}\) first

3. TRANSLATE to mathematical notation

Let A = Class A, B = Class B, L = Left-handed

• \(\mathrm{P(A)} = \mathrm{P(B)} = \frac{1}{2}\) (random class selection)
• \(\mathrm{P(L|A)} = \frac{10}{24} = \frac{5}{12}\)
• \(\mathrm{P(L|B)} = \frac{9}{30} = \frac{3}{10}\)
• Want: \(\mathrm{P(B|L)}\)

4. INFER we need total probability first

Using law of total probability:

\(\mathrm{P(L)} = \mathrm{P(L|A)} \times \mathrm{P(A)} + \mathrm{P(L|B)} \times \mathrm{P(B)}\)

\(\mathrm{P(L)} = \left(\frac{5}{12}\right) \times \left(\frac{1}{2}\right) + \left(\frac{3}{10}\right) \times \left(\frac{1}{2}\right) = \frac{5}{24} + \frac{3}{20}\)

5. SIMPLIFY the fraction addition

• Need common denominator for \(\frac{5}{24} + \frac{3}{20}\)
• LCD = 120: \(\frac{5}{24} = \frac{25}{120}\) and \(\frac{3}{20} = \frac{18}{120}\)
• \(\mathrm{P(L)} = \frac{25}{120} + \frac{18}{120} = \frac{43}{120}\)

6. INFER and apply Bayes' theorem

\(\mathrm{P(B|L)} = \frac{\mathrm{P(L|B)} \times \mathrm{P(B)}}{\mathrm{P(L)}}\)

\(\mathrm{P(B|L)} = \frac{\left(\frac{3}{10}\right) \times \left(\frac{1}{2}\right)}{\left(\frac{43}{120}\right)} = \frac{\left(\frac{3}{20}\right)}{\left(\frac{43}{120}\right)}\)

7. SIMPLIFY the complex fraction

\(\mathrm{P(B|L)} = \left(\frac{3}{20}\right) \times \left(\frac{120}{43}\right) = \frac{3 \times 120}{20 \times 43} = \frac{360}{860} = \frac{18}{43}\)

Answer: D) 18/43

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students confuse \(\mathrm{P(B|L)}\) with \(\mathrm{P(L|B)}\) and simply calculate the left-handed rate in Class B: \(\frac{9}{30} = \frac{3}{10}\).

This is the classic Bayes' confusion - they read "given the student is left-handed, what's the probability they're from Class B" but calculate "given the student is from Class B, what's the probability they're left-handed." The conditioning goes the wrong direction! This may lead them to select Choice B (3/10).

Second Most Common Error:

Missing conceptual knowledge: Students recognize they need conditional probability but forget to account for the different class sizes and equal selection probabilities. They might incorrectly think the answer is simply \(\frac{9}{9+10} = \frac{9}{19}\), treating this like drawing from a combined pool.

This may lead them to select Choice E (9/19).

The Bottom Line:

This problem tests whether students truly understand conditional probability direction and can properly apply Bayes' theorem. The key insight is that INFERring the need to "flip" the conditional probability is what separates successful students from those who get trapped by the more intuitive (but wrong) direct calculation.