| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Finding unknown probability from total probability |
| Difficulty | Moderate -0.5 This is a straightforward application of the law of total probability to find an unknown probability, followed by a basic independence calculation. Part (a) requires setting up one equation with clear given probabilities (no complex tree needed), and part (b) is direct multiplication. Slightly easier than average due to the routine nature of both parts. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.6 \times 0.7 + 0.4(1-x) = 0.58 \equiv 0.42 + 0.4(1-x) = 0.58\) | M1 | Equation of form \(0.6 \times a + 0.4 \times b = 0.58\); \(a = 0.3, 0.7\), \(b = x, (1-x)\) |
| B1 | Single correct product seen, condone \(0.42\), in an equation of appropriate form | |
| \(x = 0.6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.6 \times 0.3 + 0.4x = 0.42 \equiv 0.18 + 0.4x = 0.42\) | M1 | Equation of form \(0.6 \times a + 0.4 \times b = 0.42\); \(a = 0.3, 0.7\), \(b = x, (1-x)\) |
| B1 | Single correct product seen, condone \(0.18\), in an equation of appropriate form | |
| \(x = 0.6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0.6 \times 0.3)^2\) | M1 | \((a \times b)^2\), \(a = 0.6, 0.4\) and \(b = 0.7, 0.3, x, (1-x)\) or \(0.18^2\), alone |
| \(0.0324\) | A1 |
## Question 2:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.6 \times 0.7 + 0.4(1-x) = 0.58 \equiv 0.42 + 0.4(1-x) = 0.58$ | M1 | Equation of form $0.6 \times a + 0.4 \times b = 0.58$; $a = 0.3, 0.7$, $b = x, (1-x)$ |
| | B1 | Single correct product seen, condone $0.42$, in an equation of appropriate form |
| $x = 0.6$ | A1 | |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.6 \times 0.3 + 0.4x = 0.42 \equiv 0.18 + 0.4x = 0.42$ | M1 | Equation of form $0.6 \times a + 0.4 \times b = 0.42$; $a = 0.3, 0.7$, $b = x, (1-x)$ |
| | B1 | Single correct product seen, condone $0.18$, in an equation of appropriate form |
| $x = 0.6$ | A1 | |
**Total: 3 marks**
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.6 \times 0.3)^2$ | M1 | $(a \times b)^2$, $a = 0.6, 0.4$ and $b = 0.7, 0.3, x, (1-x)$ or $0.18^2$, alone |
| $0.0324$ | A1 | |
**Total: 2 marks**
---2 The probability that a student at a large music college plays in the band is 0.6. For a student who plays in the band, the probability that she also sings in the choir is 0.3 . For a student who does not play in the band, the probability that she sings in the choir is $x$. The probability that a randomly chosen student from the college does not sing in the choir is 0.58 .
\begin{enumerate}[label=(\alph*)]
\item Find the value of $x$.\\
Two students from the college are chosen at random.
\item Find the probability that both students play in the band and both sing in the choir.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q2 [5]}}