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Probabilistic reasoning questions appear in approximately 2–4 of the 29 Decision Making questions. They test your ability to calculate basic probabilities, understand complementary events, and apply logical reasoning to uncertain situations. This lesson covers the fundamental probability concepts you need, with a focus on the specific types of problems that appear in the UCAT.
Probability is a measure of how likely an event is to occur, expressed as a number between 0 and 1 (or equivalently, between 0% and 100%).
| Probability | Meaning |
|---|---|
| 0 | Impossible — the event cannot occur |
| 0.5 | Equally likely to occur or not occur |
| 1 | Certain — the event must occur |
P(event)=total number of possible outcomesnumber of favourable outcomes
This formula applies when all outcomes are equally likely.
Example: A bag contains 3 red balls and 7 blue balls. What is the probability of picking a red ball at random?
P(red)=3+73=103=0.3
The complement of an event A is "not A" — the event that A does not occur. The probabilities of an event and its complement always sum to 1.
P(A)+P(not A)=1
Therefore:
P(not A)=1−P(A)
The complement rule is particularly useful for "at least one" problems, which are common in UCAT DM.
Example: A medical test has a 10% chance of producing a false positive. If three independent patients are tested, what is the probability that at least one receives a false positive?
Direct approach (complicated): Calculate P(exactly 1 false positive) + P(exactly 2) + P(exactly 3). This requires multiple calculations.
Complement approach (much simpler):
P(at least one false positive)=1−P(no false positives)
P(no false positives)=0.9×0.9×0.9=0.729
P(at least one false positive)=1−0.729=0.271
UCAT Tip: Whenever you see "at least one," immediately think of the complement approach. It is almost always faster.
These problems follow a standard pattern:
| Step | Action |
|---|---|
| 1 | Identify the probability of the event NOT happening in a single trial |
| 2 | Raise this to the power of the number of trials (for independent events) |
| 3 | Subtract from 1 |
Example: The probability that a randomly selected applicant meets the UCAT threshold is 0.6. If three applicants are selected independently, what is the probability that at least one meets the threshold?
UCAT questions may express probabilities as fractions, decimals, or percentages. You should be comfortable converting between all three.
| Fraction | Decimal | Percentage |
|---|---|---|
| 1/4 | 0.25 | 25% |
| 1/3 | 0.333... | 33.3% |
| 1/2 | 0.5 | 50% |
| 2/3 | 0.667... | 66.7% |
| 3/4 | 0.75 | 75% |
| 1/5 | 0.2 | 20% |
| 1/10 | 0.1 | 10% |
Speed Tip: Memorise the common fraction-to-decimal conversions above. In the UCAT, spending 10 seconds on a conversion is 10 seconds lost from your solving time.
Many UCAT probability questions involve scenarios where outcomes are equally likely — drawing names from a hat, selecting at random, rolling a fair die.
Example: A committee of 12 members includes 5 doctors and 7 nurses. One member is selected at random to chair the meeting. What is the probability that the chair is a doctor?
P(doctor)=125
If outcomes are not equally likely, you cannot simply count favourable outcomes. Instead, you must use the given probabilities directly.
Example: The probability that a patient arriving at A&E needs surgery is 0.15. The probability that they need only observation is 0.45. The probability that they need medication but not surgery is 0.40.
These probabilities are given, not calculated from equally likely outcomes. They sum to 1 (0.15 + 0.45 + 0.40 = 1.00), confirming they cover all possibilities.
Two events are mutually exclusive if they cannot both occur at the same time.
Example: Rolling a 3 and rolling a 5 on a single die roll are mutually exclusive — you cannot roll both simultaneously.
For mutually exclusive events:
P(A or B)=P(A)+P(B)
Warning: This addition rule ONLY works for mutually exclusive events. If events can overlap, you must subtract the overlap (covered in the next lesson).
Example: A hospital has three wards. The probability a patient is assigned to Ward A is 0.3, Ward B is 0.5, and Ward C is 0.2. What is the probability a patient is assigned to Ward A or Ward C?
Since a patient goes to exactly one ward (mutually exclusive):
P(A or C)=0.3+0.2=0.5
A bag contains 4 red tokens, 6 blue tokens, and 2 green tokens. One token is drawn at random. What is the probability that the token is NOT blue?
Step 1: Total tokens = 4 + 6 + 2 = 12.
Step 2: P(blue) = 6/12 = 1/2.
Step 3: P(not blue) = 1 − 1/2 = 1/2.
Alternatively: P(not blue) = P(red or green) = (4 + 2)/12 = 6/12 = 1/2. Same answer.
| Error | Example | Correction |
|---|---|---|
| Forgetting the total | "4 out of 6 red" instead of "4 out of 12" | Always calculate the total of ALL items first |
| Adding non-mutually-exclusive events | Adding P(A) + P(B) when A and B can overlap | Check whether events can occur simultaneously |
| Confusing "at least one" with "exactly one" | Calculating P(exactly 1) when asked for P(at least 1) | "At least one" includes exactly 1, exactly 2, etc. — use the complement |
| Using wrong denominator | Using only the favourable group as denominator | The denominator is always the TOTAL number of possible outcomes |