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1
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\(\dfrac{12}{25}\)
Worked solution
\(P(RB) + P(BR) = \frac{4}{10}\cdot\frac{6}{9} + \frac{6}{10}\cdot\frac{4}{9} = \frac{48}{90} = \frac{8}{15}\).
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.
Answer D · Probability (M7.7) · question PRO-01
2
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\(\dfrac{1}{3}\)
Worked solution
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.
Answer C · Probability (M7.6) · question PRO-02
3
\(P(A) = 0.6\), \(P(B) = 0.5\) 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
Worked solution
\(P(A)P(B) = 0.3 = P(A \text{ and } B)\), 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.
Answer F · Probability (M7.7) · question PRO-03
4
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\(20\)
Worked solution
\(0.15 \times 200 = 30\).
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 \(= n \times p\).
Answer A · Probability (M7.2) · question PRO-05
5
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\(\dfrac{5}{12}\)
Worked solution
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.
Answer C · Probability (M7.7) · question PRO-06
6
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\(\dfrac{1}{2}\)
Worked solution
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.
Answer F · Probability (M7.6) · question PRO-07
7
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\(0.3\)
Worked solution
\(0.1 + 3x = 1 \Rightarrow x = 0.3\); 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'.
Answer D · Probability (M7.4) · question PRO-09
8
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\(\dfrac{91}{216}\)
Worked solution
\(\left(\frac56\right)^2 \cdot \frac16 = \frac{25}{216}\).
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.
Answer A · Probability (M7.7) · question PRO-10
9
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\(\dfrac{4}{52} \times \dfrac{4}{52}\)
Worked solution
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.
Answer E · Probability (M7.6) · question PRO-11
10
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\(0.42\)
Worked solution
\(P(A \text{ or } B) = 0.7\), 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.
Answer A · Probability (M7.7) · question PRO-12
11
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
Worked solution
\(2p(1 - p) = \frac38 \Rightarrow 16p^2 - 16p + 3 = 0 \Rightarrow (4p - 1)(4p - 3) = 0\).
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.
Answer D · Probability (M7.7) · question PRO-04
12
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\(9\)
Worked solution
\(\dfrac{n}{n+3}\cdot\dfrac{n-1}{n+2} = \dfrac{5}{12} \Rightarrow 12n^2 - 12n = 5n^2 + 25n + 30 \Rightarrow 7n^2 - 37n - 30 = 0 \Rightarrow (7n + 5)(n - 6) = 0\), 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.
Answer A · Probability (M7.7) · question PRO-08
13
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.
Worked solution
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.
Answer C · Probability (M7.7) · question X2-19
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