| 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.\\
(i) Find the probability that both balls chosen are red.\\
(ii) Show that the probability that the balls chosen are of different colours is $\frac { 15 } { 28 }$.\\
(iii) 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.\\
(iv) Draw up the probability distribution table for $X$.\\
(v) Find $\operatorname { Var } ( X )$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q6 [10]}}