| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Two-way table probabilities |
| Difficulty | Easy -1.2 This is a straightforward two-way table probability question requiring only basic probability calculations: simple probability from totals, independence check using P(A∩B)=P(A)P(B), conditional probability, and a standard combination calculation. All techniques are routine S1 content with no problem-solving insight needed. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Art | Music | Drama | |
| Boys | 24 | 40 | 32 |
| Girls | 15 | 12 | 37 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(52/160 = 13/40,\ 0.325\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{boy}) = 96/160\); \(P(\text{Music}) = 52/160\); \(P(\text{boy and Music}) = 40/160\) | M1 | Use of \(P(B) \times P(M) = P(B \cap M)\), appropriate probabilities used |
| \(96/160 \times 52/160 \neq 40/160\): Not independent | A1 | Numerical comparison and conclusion stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{not Music}\mid\text{girl}) = \frac{P(\text{not Music and girl})}{P(\text{girl})} = \frac{27/160}{64/160}\) | M1 | Appropriate probabilities in a fraction |
| \(= \dfrac{27}{64}\) | A1 | Correct answer www implies method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Direct from table | M1 | \(27/a\) or \(b/64\), \(a \neq 160\) |
| \(\dfrac{27}{64}\) | A1 | Correct answer www implies method |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(BM) \times P(BNM) \times P(GNM)\) or \(P(GM) \times P(BNM) \times P(BNM)\) | M1 | One scenario identified with 3 probs multiplied |
| \(\frac{40}{160} \times \frac{56}{159} \times \frac{52}{158}\) or \(\frac{12}{160} \times \frac{56}{159} \times \frac{55}{158}\) | A1 | One scenario correct (ignore multiplying factor) |
| \(\times 3!\) \(\times 3!/2!\) | B1 | Both multiplying factors correct |
| \(0.17387 \quad 0.02759\) \(P = 0.17387 + 0.02759\) | M1 | Both cases attempted and added (multiplying factor not required), accept unsimplified |
| \(= 0.201\) | A1 | Correct answer, oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{\dbinom{40}{1}\times\dbinom{56}{1}\times\dbinom{52}{1}+\dbinom{12}{1}\times\dbinom{56}{2}}{\dbinom{160}{3}}\) | M1 | One scenario identified with 2 or 3 combinations multiplied |
| A1 | One scenario correct | |
| B1 | Denominator correct | |
| \(\dfrac{116480 + 18480}{669920}\) | M1 | Both scenarios attempted and added, seen as numerator of a fraction |
| \(\dfrac{1687}{8374}\) | A1 | Correct answer, oe |
| 5 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $52/160 = 13/40,\ 0.325$ | B1 | oe |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{boy}) = 96/160$; $P(\text{Music}) = 52/160$; $P(\text{boy and Music}) = 40/160$ | M1 | Use of $P(B) \times P(M) = P(B \cap M)$, appropriate probabilities used |
| $96/160 \times 52/160 \neq 40/160$: Not independent | A1 | Numerical comparison and conclusion stated |
## Question 7(iii):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{not Music}\mid\text{girl}) = \frac{P(\text{not Music and girl})}{P(\text{girl})} = \frac{27/160}{64/160}$ | M1 | Appropriate probabilities in a fraction |
| $= \dfrac{27}{64}$ | A1 | Correct answer www implies method |
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Direct from table | M1 | $27/a$ or $b/64$, $a \neq 160$ |
| $\dfrac{27}{64}$ | A1 | Correct answer www implies method |
| | **2** | |
---
## Question 7(iv):
**Method 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(BM) \times P(BNM) \times P(GNM)$ **or** $P(GM) \times P(BNM) \times P(BNM)$ | M1 | One scenario identified with 3 probs multiplied |
| $\frac{40}{160} \times \frac{56}{159} \times \frac{52}{158}$ **or** $\frac{12}{160} \times \frac{56}{159} \times \frac{55}{158}$ | A1 | One scenario correct (ignore multiplying factor) |
| $\times 3!$ $\times 3!/2!$ | B1 | Both multiplying factors correct |
| $0.17387 \quad 0.02759$ $P = 0.17387 + 0.02759$ | M1 | Both cases attempted and added (multiplying factor not required), accept unsimplified |
| $= 0.201$ | A1 | Correct answer, oe |
*Note: If score is 0, award SCB1 for $\dfrac{1}{160} \times \dfrac{1}{159} \times \dfrac{1}{158} \times k$, for positive integer $k$, seen*
**Method 2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{\dbinom{40}{1}\times\dbinom{56}{1}\times\dbinom{52}{1}+\dbinom{12}{1}\times\dbinom{56}{2}}{\dbinom{160}{3}}$ | M1 | One scenario identified with 2 or 3 combinations multiplied |
| | A1 | One scenario correct |
| | B1 | Denominator correct |
| $\dfrac{116480 + 18480}{669920}$ | M1 | Both scenarios attempted and added, seen as numerator of a fraction |
| $\dfrac{1687}{8374}$ | A1 | Correct answer, oe |
| | **5** | |7 In a group of students, the numbers of boys and girls studying Art, Music and Drama are given in the following table. Each of these 160 students is studying exactly one of these subjects.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Art & Music & Drama \\
\hline
Boys & 24 & 40 & 32 \\
\hline
Girls & 15 & 12 & 37 \\
\hline
\end{tabular}
\end{center}
(i) Find the probability that a randomly chosen student is studying Music.\\
(ii) Determine whether the events 'a randomly chosen student is a boy' and 'a randomly chosen student is studying Music' are independent, justifying your answer.\\
(iii) Find the probability that a randomly chosen student is not studying Drama, given that the student is a girl.\\
(iv) Three students are chosen at random. Find the probability that exactly 1 is studying Music and exactly 2 are boys.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE S1 2018 Q7 [10]}}