Combined Probabilities

Many UCAT Decision Making probability questions involve combining the probabilities of two or more events. This lesson covers the AND rule (multiplication), the OR rule (addition), the critical distinction between independent and dependent events, and the use of tree diagrams for systematic probability calculation.

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The AND Rule (Multiplication)

When you need the probability that two events both occur, you multiply their probabilities.

For Independent Events

Two events are independent if the occurrence of one does not affect the probability of the other.

$P(A \text{ and } B) = P(A) \times P(B)$P(A and B)=P(A)×P(B)

Example: The probability of rain on Monday is 0.3, and the probability of rain on Tuesday is 0.4. If these are independent, the probability of rain on both days is:

$P(\text{rain Monday and Tuesday}) = 0.3 \times 0.4 = 0.12$P(rain Monday and Tuesday)=0.3×0.4=0.12

For Dependent Events

Two events are dependent if the occurrence of one changes the probability of the other.

$P(A \text{ and } B) = P(A) \times P(B \mid A)$P(A and B)=P(A)×P(B∣A)

Where P(B | A) means "the probability of B given that A has occurred."

Example: A bag contains 5 red and 3 blue balls. You draw two balls without replacement. What is the probability both are red?

$P(\text{1st red}) = \frac{5}{8}$P(1st red)=85​

$P(\text{2nd red} \mid \text{1st red}) = \frac{4}{7}$P(2nd red∣1st red)=74​

$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$P(both red)=85​×74​=5620​=145​

UCAT Tip: "Without replacement" signals dependent events. "With replacement" (or separate trials) signals independent events. The question will usually make this clear — read carefully.

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The OR Rule (Addition)

When you need the probability that at least one of two events occurs (A or B or both), you add their probabilities — but you must account for any overlap.

For Mutually Exclusive Events

$P(A \text{ or } B) = P(A) + P(B)$P(A or B)=P(A)+P(B)

For Non-Mutually Exclusive Events

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$P(A or B)=P(A)+P(B)−P(A and B)

Example: In a group of 100 patients, 40 have hypertension, 30 have diabetes, and 10 have both. What is the probability a randomly selected patient has hypertension or diabetes (or both)?

$P(\text{hypertension or diabetes}) = \frac{40}{100} + \frac{30}{100} - \frac{10}{100} = \frac{60}{100} = 0.6$P(hypertension or diabetes)=10040​+10030​−10010​=10060​=0.6

Key Point: The subtraction of P(A and B) prevents double-counting the patients who have both conditions.

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Tree Diagrams

Tree diagrams are the most reliable method for solving multi-stage probability problems in the UCAT. They lay out all possible outcomes and their probabilities visually.

How to Build a Tree Diagram

1. First branch: List all possible outcomes of the first event, with their probabilities.
2. Second branch: From each first-event outcome, list all possible outcomes of the second event, with their conditional probabilities.
3. Calculate: Multiply along each path to find the probability of that specific sequence.
4. Add the probabilities of all paths that match the desired outcome.

Worked Example

A diagnostic test has an 80% sensitivity (correctly identifies a positive case) and a 90% specificity (correctly identifies a negative case). In a population, 5% of people have the disease. A person is selected at random and tested. What is the probability of a positive test result?

Step 1: First branch — disease status.

• P(disease) = 0.05
• P(no disease) = 0.95

Step 2: Second branch — test result.

From "disease":

• P(positive | disease) = 0.80 (sensitivity)
• P(negative | disease) = 0.20

From "no disease":

• P(positive | no disease) = 0.10 (1 − specificity = false positive rate)
• P(negative | no disease) = 0.90

Step 3: Calculate each path.