| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Probability distribution from conditional setup |
| Difficulty | Moderate -0.3 This is a standard conditional probability question with straightforward tree diagram calculations. Parts (i)-(iii) involve basic probability rules without replacement, part (iv) requires organizing probabilities into a table, and part (v) is routine variance calculation. While multi-part, each step uses well-practiced S1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(RR) = \frac{3}{8} \times \frac{2}{7} = \frac{3}{28}\) | B1 | OE |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(RW) + P(WR)\): \(\frac{3}{8}\times\frac{5}{7}+\frac{5}{8}\times\frac{3}{7}\) | M1 | Method shown, numerical calculations identified, may include replacements |
| \(= \frac{15}{28}\) | A1 | AG, Fully correct calculations |
| Alternative: \(1-(P(RR)+P(WW))\): \(1-\left(\frac{3}{28}+\frac{5}{8}\times\frac{4}{7}\right)\) | M1 | Method shown, numerical calculations identified, may include replacements |
| \(= \frac{15}{28}\) | A1 | AG, Fully correct calculations |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{first red} \mid \text{second red}) = \frac{\textit{their (i)}}{\textit{their (i)}+\frac{5}{8}\times\frac{3}{7}} = \frac{\frac{3}{8}\times\frac{2}{7}}{\frac{3}{8}\times\frac{2}{7}+\frac{5}{8}\times\frac{3}{7}} = \frac{\frac{3}{28}}{\frac{21}{56}}\) | M1 | Conditional probability formula used consistent with *their* probabilities or correct |
| \(= \frac{2}{7}\) | A1 | OE |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): \(0\), \(1\), \(2\); \(p\): \(\frac{10}{28}\), \(\frac{15}{28}\), \(\frac{3}{28}\) | B1 | Probability distribution table with correct values of \(x\) and at least one correct probability placed. Extra \(x\) values allowed with probability of zero stated |
| B1FT | Fully correct. FT \(P(2) = \textit{their (i)}\), \(P(1) = \textit{their (ii)}\), \(\Sigma(p) = 1\) | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \frac{30}{56}+\frac{12}{56} = \frac{42}{56}\left(=\frac{3}{4}\right)\) | B1 | May be implied by use in variance formula |
| \(\text{Var}(X) = \frac{30}{56}+\frac{24}{56}-\left(\textit{their } \frac{3}{4}\right)^2\) | M1 | Substitute into correct variance formula, must have \('-\textit{their mean}^2\)'. Must be for 2 or more non-zero \(x\)-values |
| \(\frac{45}{112}\) or \(0.402\) | A1 | Correct final answer |
| 3 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(RR) = \frac{3}{8} \times \frac{2}{7} = \frac{3}{28}$ | B1 | OE |
| | **1** | |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(RW) + P(WR)$: $\frac{3}{8}\times\frac{5}{7}+\frac{5}{8}\times\frac{3}{7}$ | M1 | Method shown, numerical calculations identified, may include replacements |
| $= \frac{15}{28}$ | A1 | AG, Fully correct calculations |
| **Alternative:** $1-(P(RR)+P(WW))$: $1-\left(\frac{3}{28}+\frac{5}{8}\times\frac{4}{7}\right)$ | M1 | Method shown, numerical calculations identified, may include replacements |
| $= \frac{15}{28}$ | A1 | AG, Fully correct calculations |
| | **2** | |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{first red} \mid \text{second red}) = \frac{\textit{their (i)}}{\textit{their (i)}+\frac{5}{8}\times\frac{3}{7}} = \frac{\frac{3}{8}\times\frac{2}{7}}{\frac{3}{8}\times\frac{2}{7}+\frac{5}{8}\times\frac{3}{7}} = \frac{\frac{3}{28}}{\frac{21}{56}}$ | M1 | Conditional probability formula used consistent with *their* probabilities or correct |
| $= \frac{2}{7}$ | A1 | OE |
| | **2** | |
---
## Question 6(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: $0$, $1$, $2$; $p$: $\frac{10}{28}$, $\frac{15}{28}$, $\frac{3}{28}$ | B1 | Probability distribution table with correct values of $x$ and at least one correct probability placed. Extra $x$ values allowed with probability of zero stated |
| | B1FT | Fully correct. FT $P(2) = \textit{their (i)}$, $P(1) = \textit{their (ii)}$, $\Sigma(p) = 1$ |
| | **2** | |
---
## Question 6(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \frac{30}{56}+\frac{12}{56} = \frac{42}{56}\left(=\frac{3}{4}\right)$ | B1 | May be implied by use in variance formula |
| $\text{Var}(X) = \frac{30}{56}+\frac{24}{56}-\left(\textit{their } \frac{3}{4}\right)^2$ | M1 | Substitute into correct variance formula, must have $'-\textit{their mean}^2$'. Must be for 2 or more non-zero $x$-values |
| $\frac{45}{112}$ or $0.402$ | A1 | Correct final answer |
| | **3** | |
---6 A box contains 3 red balls and 5 white balls. One ball is chosen at random from the box and is not returned to the box. A second ball is now chosen at random from the box.\\
\begin{enumerate}[label=(\roman*)]
\item Find the probability that both balls chosen are red.
\item Show that the probability that the balls chosen are of different colours is $\frac { 15 } { 28 }$.
\item Given that the second ball chosen is red, find the probability that the first ball chosen is red.\\
The random variable $X$ denotes the number of red balls chosen.
\item Draw up the probability distribution table for $X$.
\item Find $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2019 Q6 [10]}}