UCAT Decision Making Probabilistic Reasoning: Probability Rules and Worked Examples

At TheUKCATPeople, I am Dr Akash, and probabilistic reasoning is the one Decision Making type that asks for real calculation. Most students last met probability at GCSE, so a quick, correct refresher plus a clear rule for which method to use turns these into dependable marks. This guide covers every rule the UCAT tests, with worked examples and the traps that catch people out.

In the 2026 UCAT, Decision Making is 35 questions in 37 minutes, just over a minute per question, with a simple on-screen calculator and no negative marking (UCAT Consortium).

The UCAT Consortium describes Decision Making by its answer formats rather than publishing a fixed list of named question types. The type names used across preparation, including this one, come from the official UCAT question tutorials and practice materials.

The UCAT states that no specialist mathematical or logical terminology is required. You will not be asked to quote formulae, but you do need to apply GCSE-level probability quickly. Decision Making is scored from 300 to 900, like the other cognitive subtests.

This guide sits under the Decision Making complete guide. Because it involves arithmetic, the mental methods in the Quantitative Reasoning complete guide are worth having to hand.

Probability basics

A probability is a number from 0, meaning impossible, to 1, meaning certain, and you can write it as a fraction, a decimal or a percentage. One quarter, 0.25 and 25% are the same probability. The basic rule is: probability = favourable outcomes divided by the total number of equally likely outcomes.

The five rules the UCAT tests

The AND rule: multiply for independent events

When you want two independent events to both happen, multiply their probabilities. Independent means one outcome does not change the other, such as two separate machines or two coin tosses. P(A and B) = P(A) x P(B).

Worked example 1: the AND rule

A ward has two independent alarm systems. In any given hour, alarm A fails with probability 0.05 and alarm B fails with probability 0.04. What is the probability that both fail in the same hour?

A) 0.002

B) 0.09

C) 0.20

D) 0.0002

Working:

• Both happening means multiply: 0.05 x 0.04.
• 0.05 x 0.04 = 0.002.

Answer: A) 0.002

Trap B) 0.09 adds the two probabilities. Add is for "either", multiply is for "both".

The OR rule: add for mutually exclusive events

When you want either of two events that cannot both happen at once, add their probabilities. Mutually exclusive means they cannot occur together, such as rolling a 2 or a 5 on one die. P(A or B) = P(A) + P(B). If the events can overlap, subtract the overlap: P(A) + P(B) − P(A and B).

Worked example 2: the OR rule

A fair six-sided die is rolled once. What is the probability of rolling a 2 or a 5?

A) 1/3

B) 1/6

C) 1/12

D) 5/6

Working:

• Rolling a 2 and rolling a 5 cannot both happen on one roll, so add.
• 1/6 + 1/6 = 2/6 = 1/3.

Answer: A) 1/3

Trap B) 1/6 gives the probability of just one of the two outcomes.

Worked example 2b: the OR rule when events can overlap

In a clinic, the probability a patient has diabetes is 0.3, the probability they have hypertension is 0.4, and the probability they have both is 0.15. What is the probability a patient has at least one of the two conditions?

A) 0.70

B) 0.55

C) 0.85

D) 0.15

Working:

• The two conditions can occur together, so use P(A) + P(B) − P(A and B).
• 0.3 + 0.4 − 0.15 = 0.55.

Answer: B) 0.55

Trap A) 0.70 adds 0.3 and 0.4 without subtracting the overlap, which double-counts the patients who have both. The Venn diagram shows why: the 15 with both are inside both circles.

The complement rule: one minus, for "at least one"

The probability an event does not happen is 1 minus the probability it does. This is the fastest route to "at least one" questions: P(at least one) = 1 − P(none).

Worked example 3: the complement rule

A diagnostic test is run three times, independently. Each run fails to detect a present condition with probability 0.2. What is the probability that at least one of the three runs detects it?

A) 0.8

B) 0.992

C) 0.6

D) 0.008

Working:

• P(all three miss) = 0.2 x 0.2 x 0.2 = 0.008.
• P(at least one detects) = 1 − 0.008 = 0.992.

Answer: B) 0.992

Trap D) 0.008 is the probability that all three miss. Going via the complement is far faster than adding every "one, two or three detect" case.

Odds and probability: convert both ways

Odds are stated as favourable to unfavourable. If the odds are a to b, the probability is a divided by (a + b). So odds of 1 to 4 give a probability of 1/5, which is 0.2, not 1/4.

Worked example 4: odds to probability

A treatment's success is quoted as odds of 3 to 2 in favour. What is the probability of success?

A) 0.6

B) 1.5

C) 0.4

D) 0.67

Working:

• Probability = favourable / (favourable + unfavourable) = 3 / (3 + 2).
• 3 / 5 = 0.6.

Answer: A) 0.6

Trap B) 1.5 is 3 divided by 2, treating odds as a fraction. Odds compare success to failure; probability compares success to the total.

Expected frequency: probability times trials

The expected number of times an event happens is its probability multiplied by the number of trials. This also works in reverse: a probability can be read off a frequency table.

Worked example 5: expected frequency

A screening test is given to 2,000 patients. The probability that any one patient tests positive is 0.015. How many positive results are expected?

A) 30

B) 300

C) 3

D) 15

Working:

• Expected number = probability x number of trials.
• 0.015 x 2,000 = 30.

Answer: A) 30

Dependent events: when the odds change

If an event changes what is left, the second probability is conditional on the first. This is called conditional probability, and it happens whenever you take items without replacement.

Worked example 6: without replacement

A box holds 4 red and 6 blue capsules, 10 in total. Two capsules are drawn at random without replacement. What is the probability that both are red?

A) 0.16

B) 0.133

C) 0.4

D) 0.2

Working:

• First red: 4 out of 10.
• After one red is removed, second red: 3 out of 9.
• 4/10 x 3/9 = 12/90 = 2/15 = 0.133.

Answer: B) 0.133

Trap A) 0.16 is (4/10) x (4/10), which would be correct only if the first capsule were replaced. Without replacement, the denominator drops.

Which rule do I use? A ten-second check

The traps that cost marks

• Adding when you should multiply, or the reverse. Both means multiply; either means add.
• Confusing odds with probability: odds of 1 to 4 is a probability of 1/5, not 1/4.
• Treating independent and mutually exclusive as the same thing. They are opposites: independent events can both happen, mutually exclusive events cannot.
• Forgetting to change the denominator in without-replacement problems.
• The gambler’s fallacy: past independent results do not change the next one. After ten heads, the next toss is still one half.
• Answering "at least one" the long way instead of using 1 minus the probability of none.

Keep to about a minute per question using the timings guide, and use the simple on-screen calculator only for awkward multiplications, since reaching for it costs a few seconds each time.

Once the rules are second nature, put them under time pressure in our data logic practice questions and our free DM practice questions.

Test yourself

Work these before you read the answers below the list.

1. Two independent components each work with probability 0.9. What is the probability that both work?

2. A test is run twice, independently, and each run misses a condition with probability 0.1. What is the probability that at least one run detects it?

3. A treatment has odds of 4 to 1 in favour of success. What is the probability of success?

Answers

1. 0.81. Both working means multiply: 0.9 x 0.9 = 0.81.

2. 0.99. P(both miss) = 0.1 x 0.1 = 0.01, so P(at least one detects) = 1 − 0.01 = 0.99.

3. 0.8. Probability = 4 divided by (4 + 1) = 4/5 = 0.8.

Key Takeaway: Both means multiply, either means add, at least one means one minus the probability of none, and odds of a to b give a probability of a over a plus b. Keep independent and mutually exclusive events apart, drop the denominator when there is no replacement, and sanity-check that every answer sits between 0 and 1.