Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.
1. PRO-01
A bag contains \(4\) red and \(6\) blue counters. Two counters are taken at random without replacement. Find the probability that they are different colours.
- A\(\dfrac{24}{100}\)
- B\(\dfrac{7}{15}\)
- C\(\dfrac{2}{5}\)
- D\(\dfrac{8}{15}\)
- E\(\dfrac{4}{15}\)
- F\(\dfrac{12}{25}\)
Answer: D. \(P(RB) + P(BR) = \frac{4}{10}\cdot\frac{6}{9} + \frac{6}{10}\cdot\frac{4}{9} = \frac{48}{90} = \frac{8}{15}\).
Full method — how and why
'Different colours' happens in two orders. Either add the two ordered cases or use the complement.
- Red then blue: \(\dfrac{4}{10} \times \dfrac{6}{9}= \dfrac{24}{90}\).After removing a red, \(9\) counters remain and all \(6\) blues are still there - without replacement.
- Blue then red: \(\dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{24}{90}\).The same product, but the orders are distinct outcomes so both count.
- Add: \(\dfrac{48}{90} = \dfrac{8}{15}\).Mutually exclusive cases add; then cancel by \(6\).
- Check by complement: \(P(\text{same}) = \frac{4}{10}\frac{3}{9} + \frac{6}{10}\frac{5}{9} = \frac{12+30}{90} = \frac{42}{90}\), and \(1 - \frac{42}{90} = \frac{48}{90}\) ✓.Two routes agreeing is strong confirmation.
Where it goes wrong
- Counting only one order, giving \(\frac{4}{15}\) - an offered option.
- Keeping the denominator at \(10\) for the second draw (\(\frac{12}{25}\)), which is the with replacement answer.
Takeaway Without replacement: reduce both the relevant colour count and the total; and count both orders unless the question fixes one.
2. PRO-02
Two fair dice are thrown. Find the probability that the product of the two scores is odd.
- A\(\dfrac{2}{3}\)
- B\(\dfrac{3}{4}\)
- C\(\dfrac{1}{4}\)
- D\(\dfrac{1}{6}\)
- E\(\dfrac{1}{2}\)
- F\(\dfrac{1}{3}\)
Answer: C. The product is odd only if both scores are odd: \(\frac12 \times \frac12 = \frac14\).
Full method — how and why
A product is odd only when every factor is odd - so this is an 'and' of two independent events.
- Odd \(\times\) odd \(=\) odd; any even factor makes the product even.Recognising this converts the question into a much simpler one.
- \(P(\text{first die odd}) = \frac36 = \frac12\); same for the second.Three odd faces out of six: \(1, 3, 5\).
- Independent dice, so multiply: \(\frac12 \times \frac12 = \frac14\).One die's outcome has no effect on the other.
Where it goes wrong
- Answering \(\frac12\) by thinking 'half of all products are odd' - in fact only \(9\) of the \(36\) outcomes are.
- Treating it as a sum (where odd \(+\) odd \(=\) even), a different and much more common question.
Takeaway Product odd \(\iff\) all factors odd; product even \(\iff\) at least one even.
3. PRO-03
\(P(A) = 0.6\), \(P(B) = 0.5\) and \(P(A \text{ and } B) = 0.3\). Which of the following statements is/are true?
- I\(A\) and \(B\) are independent.
- II\(A\) and \(B\) are mutually exclusive.
- III\(P(A \text{ or } B) = 0.8\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Answer: F. \(P(A)P(B) = 0.3 = P(A \text{ and } B)\), so independent (I). Not mutually exclusive since they can both happen (II false). \(P(A \text{ or } B) = 0.6 + 0.5 - 0.3 = 0.8\) (III).
Full method — how and why
Three definitions to test: independence, mutual exclusivity and the addition rule.
- I: independence means \(P(A \cap B) = P(A)P(B)\). Here \(0.6 \times 0.5 = 0.3\), which matches. True.Independence is a numerical condition, not something you can eyeball.
- II: mutually exclusive means \(P(A \cap B) = 0\). It is \(0.3 \neq 0\). False.They overlap, so they can happen together.
- III: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.3 = 0.8\). True.Subtract the overlap or you count it twice.
- So I and III only.Match to the list.
Where it goes wrong
- Treating independent and mutually exclusive as the same thing - they are near-opposites: two events with non-zero probabilities cannot be both.
- Forgetting the \(-P(A\cap B)\) and answering \(1.1\), which is impossible for a probability.
Takeaway Independent: \(P(A\cap B) = P(A)P(B)\). Mutually exclusive: \(P(A\cap B) = 0\). Check the numbers, not the words.
4. PRO-05
A spinner has probability \(0.15\) of landing on red. It is spun \(200\) times. What is the expected number of times it lands on red?
- A\(30\)
- B\(40\)
- C\(15\)
- D\(35\)
- E\(25\)
- F\(20\)
Answer: A. \(0.15 \times 200 = 30\).
Full method — how and why
Expected number of successes in repeated independent trials is simply \(n \times p\).
- Identify \(n = 200\) spins and \(p = 0.15\).Each spin is an independent trial with the same probability.
- Expected number \(= 200 \times 0.15 = 30\).\(0.15 = \frac{15}{100}\), and \(15\%\) of \(200\) is \(30\).
Where it goes wrong
- Answering \(15\) by reading the percentage as the answer.
- Expecting the answer to be a whole number in general - expectations often are not, and that is fine.
Takeaway Expected frequency \(= n \times p\).
5. PRO-06
In a class of \(30\), \(18\) study French, \(12\) study German and \(5\) study both. A student is chosen at random. Given that the student studies French, what is the probability that they also study German?
- A\(\dfrac{1}{6}\)
- B\(\dfrac{5}{25}\)
- C\(\dfrac{5}{18}\)
- D\(\dfrac{5}{30}\)
- E\(\dfrac{13}{18}\)
- F\(\dfrac{5}{12}\)
Answer: C. Conditional on French (\(18\) students), \(5\) also study German: \(\frac{5}{18}\).
Full method — how and why
The words 'given that' signal conditional probability: the sample space shrinks to the condition.
- Restrict attention to the \(18\) French students - they are now the whole universe.'Given that the student studies French' means we only consider those students.
- Of those \(18\), the number who also study German is \(5\).The overlap is stated directly in the question.
- \(P(\text{German} \mid \text{French}) = \dfrac{5}{18}\).Equivalently \(\dfrac{P(F \cap G)}{P(F)} = \dfrac{5/30}{18/30} = \dfrac{5}{18}\) - the \(30\)s cancel.
Where it goes wrong
- Using \(\frac{5}{30}\) - the unconditional probability of studying both, ignoring the condition.
- Using \(\frac{5}{12}\), which conditions on German instead - the wrong way round.
Takeaway Conditioning replaces the denominator: divide by the size of the group you are told you are in.
6. PRO-07
Three fair coins are tossed. Find the probability of getting more heads than tails.
- A\(\dfrac{3}{4}\)
- B\(\dfrac{1}{4}\)
- C\(\dfrac{1}{8}\)
- D\(\dfrac{5}{8}\)
- E\(\dfrac{3}{8}\)
- F\(\dfrac{1}{2}\)
Answer: F. Need \(2\) or \(3\) heads: \(\frac{3 + 1}{8} = \frac12\) (by symmetry between heads and tails).
Full method — how and why
Either count the favourable outcomes out of \(8\), or exploit the symmetry between heads and tails.
- With three coins there are \(2^3 = 8\) equally likely outcomes.Each coin independently doubles the number of outcomes.
- 'More heads than tails' means \(2\) or \(3\) heads: \(\binom32 + \binom33 = 3 + 1 = 4\) outcomes.With an odd number of coins there is no tie, so every outcome has a strict majority.
- \(P = \dfrac48 = \dfrac12\).Or instantly: heads and tails are symmetric and a tie is impossible, so each must have probability \(\frac12\).
Where it goes wrong
- Answering \(\frac38\) - the probability of exactly \(2\) heads, forgetting the all-heads case.
- Trying the symmetry argument with an even number of coins, where ties break it.
Takeaway An odd number of fair coins: \(P(\text{more heads}) = \frac12\) by symmetry, since ties cannot occur.
7. PRO-09
The probabilities that a train is early, on time or late are \(0.1\), \(x\) and \(2x\) respectively. Find the probability that the train is not late.
- A\(0.5\)
- B\(0.9\)
- C\(0.7\)
- D\(0.4\)
- E\(0.6\)
- F\(0.3\)
Answer: D. \(0.1 + 3x = 1 \Rightarrow x = 0.3\); not late \(= 0.1 + 0.3 = 0.4\).
Full method — how and why
The probabilities of an exhaustive set of outcomes sum to \(1\). Use that to find \(x\) first.
- Early, on time, late are the only possibilities, so \(0.1 + x + 2x = 1\).Exhaustive and mutually exclusive outcomes have total probability \(1\).
- Solve: \(3x = 0.9\), so \(x = 0.3\) and late \(= 0.6\).Subtract \(0.1\), then divide by \(3\).
- Not late \(=\) early \(+\) on time \(= 0.1 + 0.3 = 0.4\).Or use the complement: \(1 - 0.6 = 0.4\) - quicker and a useful check.
Where it goes wrong
- Stopping at \(x = 0.3\) and answering that, or answering \(0.6\), the probability of being late.
- Forgetting 'early' counts as 'not late'.
Takeaway Probabilities of all outcomes sum to \(1\); 'not late' is the complement of 'late'.
8. PRO-10
A fair die is thrown repeatedly until a six appears. Find the probability that the first six appears on the third throw.
- A\(\dfrac{25}{216}\)
- B\(\dfrac{5}{36}\)
- C\(\dfrac{1}{216}\)
- D\(\dfrac{1}{6}\)
- E\(\dfrac{125}{216}\)
- F\(\dfrac{91}{216}\)
Answer: A. \(\left(\frac56\right)^2 \cdot \frac16 = \frac{25}{216}\).
Full method — how and why
'First success on the \(k\)th trial' means failure, failure, ..., then success - a chain of independent events.
- The first two throws must be non-sixes: probability \(\frac56\) each.'First six on the third' forbids a six on throws one and two.
- The third throw must be a six: probability \(\frac16\).Exactly one way to succeed on that throw.
- Multiply: \(\left(\frac56\right)^2 \times \frac16 = \dfrac{25}{216}\).Independent throws, so probabilities multiply.
Where it goes wrong
- Answering \(\frac16\) by ignoring the first two throws.
- Using \(\left(\frac16\right)^3 = \frac{1}{216}\), which is the probability of three sixes.
Takeaway First success on trial \(k\): \((1-p)^{k-1}p\) - failures first, then the one success.
9. PRO-11
Two cards are drawn at random from a standard pack of \(52\) without replacement. Which of the following expressions is the probability that both are aces?
- A\(\dfrac{4}{52} + \dfrac{3}{51}\)
- B\(\dfrac{2}{52}\)
- C\(\dfrac{4}{52} \times \dfrac{3}{52}\)
- D\(\dfrac{1}{13} \times \dfrac{1}{13}\)
- E\(\dfrac{4}{52} \times \dfrac{3}{51}\)
- F\(\dfrac{4}{52} \times \dfrac{4}{52}\)
Answer: E. Without replacement the second draw has \(3\) aces among \(51\) cards.
Full method — how and why
The question asks for the expression, so the reasoning is entirely about how the second draw changes.
- First card an ace: \(\dfrac{4}{52}\) - four aces in the pack.All \(52\) cards are equally likely.
- Given that, one ace and one card overall have gone: \(3\) aces remain among \(51\) cards.Without replacement, both numbers drop by one.
- Multiply: \(\dfrac{4}{52} \times \dfrac{3}{51}\).This is \(P(A) \times P(B \mid A)\), the multiplication rule for dependent events.
Where it goes wrong
- \(\frac{4}{52} \times \frac{4}{52}\) - that is with replacement.
- \(\frac{4}{52} \times \frac{3}{52}\) - reducing the aces but forgetting the total also falls to \(51\).
- Adding rather than multiplying: 'and' means multiply.
Takeaway Without replacement, reduce numerator and denominator; 'and' \(\Rightarrow\) multiply.
10. PRO-12
Events \(A\) and \(B\) are mutually exclusive with \(P(A) = 0.3\) and \(P(B) = 0.4\). Find \(P(\text{neither } A \text{ nor } B)\).
- A\(0.3\)
- B\(0.7\)
- C\(0.1\)
- D\(0.58\)
- E\(0.12\)
- F\(0.42\)
Answer: A. \(P(A \text{ or } B) = 0.7\), so neither \(= 0.3\).
Full method — how and why
'Neither' is the complement of 'at least one'. Mutual exclusivity makes the addition rule simple.
- Mutually exclusive means \(P(A \cap B) = 0\), so \(P(A \cup B) = 0.3 + 0.4 = 0.7\).No overlap to subtract.
- 'Neither' is the complement of 'A or B': \(1 - 0.7 = 0.3\).Everything not in \(A \cup B\).
Where it goes wrong
- Answering \(0.7\), the probability that one of them does happen.
- Computing \(0.7 \times 0.6 = 0.42\) by treating them as independent - they are explicitly not, and independent events with these probabilities could not be mutually exclusive.
Takeaway Neither \(= 1 - P(A \cup B)\); and for mutually exclusive events \(P(A\cup B)\) is just the sum.
11. PRO-04
A biased coin has probability \(p\) of landing heads. It is tossed twice. The probability of exactly one head is \(\dfrac{3}{8}\). Find the possible values of \(p\).
- A\(\dfrac{1}{3}\) or \(\dfrac{2}{3}\)
- B\(\dfrac{1}{2}\) only
- C\(\dfrac{1}{8}\) or \(\dfrac{7}{8}\)
- D\(\dfrac{1}{4}\) or \(\dfrac{3}{4}\)
- E\(\dfrac{3}{8}\) only
- F\(\dfrac{3}{16}\) only
Answer: D. \(2p(1 - p) = \frac38 \Rightarrow 16p^2 - 16p + 3 = 0 \Rightarrow (4p - 1)(4p - 3) = 0\).
Full method — how and why
'Exactly one head' in two tosses happens in two orders; that gives a quadratic in \(p\) with two valid roots.
- \(P(\text{exactly one head}) = P(HT) + P(TH) = p(1-p) + (1-p)p = 2p(1-p)\).The factor \(2\) is the number of orders - equivalently \(\binom21\).
- Set \(2p(1-p) = \frac38\) and clear fractions: \(16p(1-p) = 3\), so \(16p - 16p^2 = 3\).Multiply through by \(8\).
- Rearrange and factorise: \(16p^2 - 16p + 3 = 0 \Rightarrow (4p-1)(4p-3) = 0\).Check: \(4 \times 4 = 16\), \(-1 \times -3 = 3\), and \(-12p - 4p = -16p\) ✓.
- So \(p = \frac14\) or \(p = \frac34\); both lie in \([0,1]\), so both are valid.They are mirror images - swapping heads and tails leaves 'exactly one head' unchanged, so the symmetry is expected.
Where it goes wrong
- Omitting the factor \(2\), which gives \(p^2 - p + \frac38 = 0\) with no real roots.
- Discarding one root; a biased coin can favour either face, and the symmetry \(p \leftrightarrow 1-p\) guarantees a pair.
Takeaway 'Exactly one' needs the count of orders; and check both roots against \(0 \leq p \leq 1\) rather than assuming one is spurious.
12. PRO-08
A box contains \(n\) white balls and \(3\) black balls. Two balls are drawn without replacement. The probability that both are white is \(\dfrac{5}{12}\). Find \(n\).
- A\(6\)
- B\(4\)
- C\(5\)
- D\(3\)
- E\(7\)
- F\(9\)
Answer: A. \(\dfrac{n}{n+3}\cdot\dfrac{n-1}{n+2} = \dfrac{5}{12} \Rightarrow 12n^2 - 12n = 5n^2 + 25n + 30 \Rightarrow 7n^2 - 37n - 30 = 0 \Rightarrow (7n + 5)(n - 6) = 0\), so \(n = 6\).
Full method — how and why
Set up the without-replacement probability in terms of \(n\), then solve the resulting quadratic.
- There are \(n + 3\) balls in total. \(P(\text{both white}) = \dfrac{n}{n+3} \times \dfrac{n-1}{n+2}\).After one white is removed, \(n-1\) whites remain out of \(n+2\).
- Set equal to \(\frac{5}{12}\) and cross-multiply: \(12n(n-1) = 5(n+3)(n+2)\).Cross-multiplying is safe since all denominators are positive.
- Expand: \(12n^2 - 12n = 5(n^2 + 5n + 6) = 5n^2 + 25n + 30\).Expand the right side carefully: \((n+3)(n+2) = n^2 + 5n + 6\).
- Collect: \(7n^2 - 37n - 30 = 0\), which factorises as \((7n + 5)(n - 6) = 0\).Check the middle term: \(-42n + 5n = -37n\) ✓.
- \(n = -\frac57\) is impossible for a count, so \(n = 6\).Verify: \(\frac{6}{9}\times\frac58 = \frac{30}{72} = \frac{5}{12}\) ✓.
Where it goes wrong
- Writing \(\left(\frac{n}{n+3}\right)^2\), the with-replacement version.
- Accepting the negative root instead of rejecting it on contextual grounds.
Takeaway Model without replacement by decreasing both numerator and denominator, then reject roots that cannot be counts.
13. X2-19
Events \(A\) and \(B\) have \(P(A) > 0\) and \(P(B) > 0\). \(P\): '\(A\) and \(B\) are mutually exclusive'. \(Q\): '\(A\) and \(B\) are not independent'. Which one of the following is true?
- AP is necessary and sufficient for Q.
- BP is necessary but not sufficient for Q.
- CP is sufficient but not necessary for Q.
- DP is neither necessary nor sufficient for Q.
Answer: C. Mutually exclusive gives \(P(A \text{ and } B) = 0 \neq P(A)P(B)\), so they cannot be independent: \(P\) is sufficient. But two overlapping events can also fail to be independent, so \(P\) is not necessary.
Full method — how and why
Relating mutual exclusivity to independence. One implication holds; the converse has easy counterexamples.
- If \(A,B\) are mutually exclusive, \(P(A\cap B)=0\). But \(P(A)P(B)>0\) since both probabilities are positive.The positivity hypothesis is what makes the argument work.
- So \(P(A\cap B) \neq P(A)P(B)\): they are not independent. \(P \Rightarrow Q\): sufficient.Independence is exactly the equality that fails here.
- Necessity: find dependent events that are not mutually exclusive. Roll a die with \(A\): 'even' and \(B\): 'at most \(3\)'. Then \(P(A\cap B) = P(\{2\}) = \frac16\), but \(P(A)P(B) = \frac12 \times \frac12 = \frac14\).Unequal, so they are dependent - yet they overlap, so they are not mutually exclusive.
- So \(P\) is sufficient but not necessary.Match to the list.
Where it goes wrong
- Treating mutually exclusive and independent as opposites in a way that makes them necessary for each other - dependence has many other causes.
- Forgetting the hypothesis \(P(A),P(B)>0\); with a zero probability, events can be both mutually exclusive and independent.
Takeaway For events of positive probability, mutually exclusive \(\Rightarrow\) dependent, but not conversely.