| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with given score/outcome |
| Difficulty | Standard +0.3 This is a standard conditional probability question with straightforward tree diagram structure. Part (i) requires simple multiplication of probabilities, part (ii) applies Bayes' theorem in a routine context, and part (iii) involves constructing a probability distribution from already-calculated probabilities. While it requires careful bookkeeping across multiple parts, the techniques are all standard S1 material with no novel insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T,B) = \frac{5}{12} \times \frac{2}{10} = \frac{1}{12}\ (0.0833)\) | M1 | Multiply their \(P(T)\) by \(2/9\) or \(2/10\) only |
| A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(C_S \cap C_A) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120}\ (0.2333)\) | M1 | Multiply their \(P(C_S)\) by \(3/9\) or \(4/10\) seen as num or denom of a fraction |
| \(P(C_A) = \frac{7}{12}\times\frac{4}{10} + \frac{5}{12}\times\frac{3}{10} = \frac{43}{120}\ (0.3583)\) | M1 | Summing 2 two-factor products to find \(P(C_A)\) seen anywhere |
| \(P(C_S | C_A) = \frac{P(C\cap C)}{P(C_A)} = \frac{28/120}{43/120}\) | A1 |
| \(= \frac{28}{43}\ (0.651)\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 0, 1, 2\); Prob \(= 7/24,\ 19/40,\ 7/30\) | B1 | \(x = 0, 1, 2\) can be implied from table or working |
| \(P(X=0) = P(T,B) + P(T,T)\) | M1 | 1 or 2 two-factor products, denoms 12 and 10 or 12 and 9, implied if ans is correct |
| \(= \frac{5}{12}\times\frac{2}{10} + \frac{5}{12}\times\frac{5}{10} = \frac{7}{24}\ (0.292)\) | A1 | One correct unsimplified |
| \(P(X=2) = P(C,C) = \frac{7}{12}\times\frac{4}{10} = \frac{28}{120}\ (0.233)\) | B1 | One other correct unsimplified |
| \(P(X=1) = 1 - 7/24 - 28/120 = \frac{19}{40}\ (0.475)\) | B1ft [5] | Third correct ft \(1 - P(\text{2 of their probs})\) |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T,B) = \frac{5}{12} \times \frac{2}{10} = \frac{1}{12}\ (0.0833)$ | M1 | Multiply their $P(T)$ by $2/9$ or $2/10$ only |
| | A1 **[2]** | Correct answer |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(C_S \cap C_A) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120}\ (0.2333)$ | M1 | Multiply their $P(C_S)$ by $3/9$ or $4/10$ seen as num or denom of a fraction |
| $P(C_A) = \frac{7}{12}\times\frac{4}{10} + \frac{5}{12}\times\frac{3}{10} = \frac{43}{120}\ (0.3583)$ | M1 | Summing 2 two-factor products to find $P(C_A)$ seen anywhere |
| $P(C_S|C_A) = \frac{P(C\cap C)}{P(C_A)} = \frac{28/120}{43/120}$ | A1 | Correct unsimplified $P(C_A)$ seen as num or denom of a fraction |
| $= \frac{28}{43}\ (0.651)$ | A1 **[4]** | Correct answer |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 0, 1, 2$; Prob $= 7/24,\ 19/40,\ 7/30$ | B1 | $x = 0, 1, 2$ can be implied from table or working |
| $P(X=0) = P(T,B) + P(T,T)$ | M1 | 1 or 2 two-factor products, denoms 12 and 10 or 12 and 9, implied if ans is correct |
| $= \frac{5}{12}\times\frac{2}{10} + \frac{5}{12}\times\frac{5}{10} = \frac{7}{24}\ (0.292)$ | A1 | One correct unsimplified |
| $P(X=2) = P(C,C) = \frac{7}{12}\times\frac{4}{10} = \frac{28}{120}\ (0.233)$ | B1 | One other correct unsimplified |
| $P(X=1) = 1 - 7/24 - 28/120 = \frac{19}{40}\ (0.475)$ | B1ft **[5]** | Third correct ft $1 - P(\text{2 of their probs})$ |7 Susan has a bag of sweets containing 7 chocolates and 5 toffees. Ahmad has a bag of sweets containing 3 chocolates, 4 toffees and 2 boiled sweets. A sweet is taken at random from Susan's bag and put in Ahmad's bag. A sweet is then taken at random from Ahmad's bag.\\
(i) Find the probability that the two sweets taken are a toffee from Susan's bag and a boiled sweet from Ahmad's bag.\\
(ii) Given that the sweet taken from Ahmad's bag is a chocolate, find the probability that the sweet taken from Susan's bag was also a chocolate.\\
(iii) The random variable $X$ is the number of times a chocolate is taken. State the possible values of $X$ and draw up a table to show the probability distribution of $X$.
\hfill \mbox{\textit{CAIE S1 2013 Q7 [11]}}