OMPT drill — Probability & statistics

Probability basics

Sample spaces, complements, and the addition and multiplication rules. Conditional probability is where intuition fails and the formula must take over. Write the formula down before plugging in numbers; on this strand the setup is the whole battle.

Tested inOMPT-E

Lesson

When outcomes are equally likely, probability is counting twice over: the outcomes you want, divided by the outcomes there are. Rolling a die, $P (even) = \frac{3}{6} = \frac{1}{2}$ . The denominator is called the sample space, and the single most reliable habit in this topic is describing it precisely before computing anything. Two dice have 36 ordered outcomes, not 11 sums, not 21 unordered pairs, and most wrong answers here are really wrong sample spaces.

Three rules then organise everything. The complement rule, $P (not A) = 1 - P (A)$ , which turns nasty "at least one" questions into easy "none" questions. The addition rule for "or". And the multiplication rule for "and", which comes in two flavours depending on whether the first event changes the odds of the second: that distinction, independence versus dependence, is where dice-and-coins intuition stops being enough and the formulas have to take over.

Problem 1Numeric answer

Two fair dice are rolled. What is the probability that the sum equals 9? Round to two decimals.

Show the worked solution

Answer: 0.11

Out of $36$ equally likely outcomes, the sum 9 arises from $(3, 6), (4, 5), (5, 4), (6, 3)$ — four of them. So the probability is $\frac{4}{36} = \frac{1}{9} \approx 0.11$ .

Problem 2Multiple choice

Events $A$ and $B$ are independent with $P (A) = 0.4$ and $P (B) = 0.5$ . What is $P (A \cup B)$ ?

A $0.9$
B $0.7$
C $0.2$
D $0.6$

Show the worked solution

Answer: B — $0.7$

Independence gives $P (A \cap B) = 0.4 \cdot 0.5 = 0.2$ . Then the addition rule: $P (A \cup B) = 0.4 + 0.5 - 0.2 = 0.7$ .

Problem 3Numeric answer

A bag holds 5 red and 3 blue marbles. Two marbles are drawn without replacement. What is the probability both are red? Round to two decimals.

Show the worked solution

Answer: 0.36

I chain the draws: the first is red with probability $\frac{5}{8}$ ; given that, the second is red with probability $\frac{4}{7}$ . Multiplying: $\frac{5}{8} \cdot \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \approx 0.36$ .

Problem 4Multiple choice

A fair die is rolled three times. What is the probability of rolling at least one six?

A $\frac{1}{2}$
B $\frac{91}{216}$
C $\frac{1}{216}$
D $\frac{125}{216}$

Show the worked solution

Answer: B — $\frac{91}{216}$

"At least one" is easiest through the complement: the probability of no six at all is $(\frac{5}{6})^{3} = \frac{125}{216}$ . Subtracting from 1 gives $\frac{91}{216} \approx 0.42$ .

Problem 5Spot the error

In a class, 60% of students take biology and 45% take chemistry; 25% take both. A student reasons about the chance a random classmate takes biology or chemistry. Step 1: write the goal as $P (B \cup C)$ . Step 2: compute $P (B \cup C) = 0.60 + 0.45 = 1.05$ . Step 3: since probabilities cannot exceed 1, cap the answer at 1. Which step contains the error?

Show the worked solution

Answer: Step 2

Step 1 frames the problem correctly. Step 2 adds the probabilities as if the subjects were mutually exclusive, but 25% of students sit in both classes and get counted twice. The addition rule fixes it: $0.60 + 0.45 - 0.25 = 0.80$ . Step 3 is a patch over a broken calculation — getting a "probability" above 1 was the signal to go back, not to clip.

Problem 6Numeric answer

A screening test detects a condition with probability $0.95$ when present, and gives a false positive with probability $0.10$ when absent. In the tested population, 2% have the condition. Given a positive result, what is the probability the person actually has the condition? Round to two decimals.

Show the worked solution

Answer: 0.16

I build a tree over the whole population. True positives: $0.02 \cdot 0.95 = 0.019$ . False positives: $0.98 \cdot 0.10 = 0.098$ . All positives: $0.019 + 0.098 = 0.117$ . The conditional probability is $\frac{0.019}{0.117} \approx 0.16$ . Rare conditions make even good tests unreliable on a single positive — the false positives from the huge healthy group swamp the true ones.