UCAT conditional probability questions
Given that means: keep only the cases that meet the condition, then find the share. If one person who passed is picked, divide by the number who passed, not everyone. Try two below.
A course records outcomes. | Group | Passed | Did not pass | |---|---:|---:| | Morning | 31 | 11 | | Evening | 21 | 41 | One person who passed is selected at random. What is the probability that the person attended in the morning, to one decimal place?
Worked answers
1. A course records outcomes. | Group | Passed | Did not pass | |---|---:|---:| | Morning | 31 | 11 | | Evening | 21 | 41 | One person who passed is selected at random. What is the probability that the person attended in the morning, to one decimal place?
Answer: D: 59.6%
- People who passed = 31 + 21 = 52.
- Morning share among passes = 31/52 = 59.6%.
2. A batch contains 1100 components. Supplier A provides 40% and Supplier B provides 60%. Exactly 15% of A’s components and 5% of B’s components fail inspection. One failed component is selected at random. What is the probability it came from A, to one decimal place?
Answer: C: 66.7%
- Failed from A = 1100 × 0.4 × 0.15 = 66.
- Failed from B = 1100 × 0.6 × 0.05 = 33.
- A among failures = 66/(66 + 33) = 66.7%.
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Use counts, not formulas
Turn percentages into counts of a round total, like 1,000. Then the answer is a fraction of two counts.
| Step | Example: 40% from A, 15% of A fail; 60% from B, 5% of B fail |
|---|---|
| Fails from A | 1,000 × 0.4 × 0.15 = 60 |
| Fails from B | 1,000 × 0.6 × 0.05 = 30 |
| Picked failure is from A | 60 ÷ 90 = 66.7% |
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