| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.3 This is a straightforward probability question requiring systematic enumeration of outcomes with weighted coins. Part (i) guides students through one calculation, part (ii) requires completing similar calculations for all cases (0, 1, 2, 3 heads), and part (iii) is a standard expectation calculation. While it involves multiple steps and careful bookkeeping, it uses only basic probability rules (multiplication, addition) with no novel insight required, making it slightly easier than average. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(A: P(H) = \frac{2}{3},\ P(T) = \frac{1}{3}\); \(B: P(H) = \frac{1}{4},\ P(T) = \frac{3}{4}\) | M1 | Using some of \(\frac{2}{3}, \frac{1}{3}, \frac{1}{4}\) or \(\frac{3}{4}\) in a calculation involving product of 3 probabilities |
| \(P(1H) = P(HTT) + P(THT) + P(TTH)\) \(= \left(\frac{2}{3} \times \frac{1}{3} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{2}{3} \times \frac{3}{4}\right)\) | M1 | Summing 3 options not all the same |
| \(+ \left(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{4}\right) = \frac{13}{36}\) AG | A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Table with \(x\): \(0, 1, 2, 3\) and \(P\): \(\frac{3}{36}, \frac{13}{36}, \frac{16}{36}, \frac{4}{36}\) | B1 | \(0, 1, 2, 3\) seen for table; no probabilities needed, table not absolutely necessary if calculations shown |
| \(P(0H) = P(TTT) = \frac{1}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{12}\) | B1 | One probability correct other than (i); condone \(0.083\) for \(0.0833\) |
| \(P(2H) = P(HHT) + P(HTH) + P(THH)\) \(= \left(\frac{2}{3} \times \frac{2}{3} \times \frac{3}{4}\right) + \left(\frac{2}{3} \times \frac{1}{3} \times \frac{1}{4}\right) + \left(\frac{1}{3} \times \frac{2}{3} \times \frac{1}{4}\right) = \frac{4}{9}\) not \(\frac{2}{3} \times \frac{2}{3}\) | B1 | A second probability correct; need 3 factors can be implied |
| \(P(3H) = P(HHH) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{4} = \frac{1}{9}\) | B1\(\checkmark\) [4] | A third probability correct ft \(\frac{23}{36} - \Sigma\) their 2 probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(X) = \frac{13}{36} + \frac{32}{36} + \frac{12}{36}\) | M1 | Attempt to evaluate \(\Sigma xp\) at least 3 values of \(x\) in table |
| \(= \frac{57}{36}\ \left(\frac{19}{12}\right)\ (1.58)\) | A1 [2] | Correct answer |
## Question 4:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $A: P(H) = \frac{2}{3},\ P(T) = \frac{1}{3}$; $B: P(H) = \frac{1}{4},\ P(T) = \frac{3}{4}$ | M1 | Using some of $\frac{2}{3}, \frac{1}{3}, \frac{1}{4}$ or $\frac{3}{4}$ in a calculation involving product of 3 probabilities |
| $P(1H) = P(HTT) + P(THT) + P(TTH)$ $= \left(\frac{2}{3} \times \frac{1}{3} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{2}{3} \times \frac{3}{4}\right)$ | M1 | Summing 3 options not all the same |
| $+ \left(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{4}\right) = \frac{13}{36}$ **AG** | A1 **[3]** | Correct answer |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Table with $x$: $0, 1, 2, 3$ and $P$: $\frac{3}{36}, \frac{13}{36}, \frac{16}{36}, \frac{4}{36}$ | B1 | $0, 1, 2, 3$ seen for table; no probabilities needed, table not absolutely necessary if calculations shown |
| $P(0H) = P(TTT) = \frac{1}{3} \times \frac{1}{3} \times \frac{3}{4} = \frac{1}{12}$ | B1 | One probability correct other than (i); condone $0.083$ for $0.0833$ |
| $P(2H) = P(HHT) + P(HTH) + P(THH)$ $= \left(\frac{2}{3} \times \frac{2}{3} \times \frac{3}{4}\right) + \left(\frac{2}{3} \times \frac{1}{3} \times \frac{1}{4}\right) + \left(\frac{1}{3} \times \frac{2}{3} \times \frac{1}{4}\right) = \frac{4}{9}$ not $\frac{2}{3} \times \frac{2}{3}$ | B1 | A second probability correct; need 3 factors can be implied |
| $P(3H) = P(HHH) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{4} = \frac{1}{9}$ | B1$\checkmark$ **[4]** | A third probability correct ft $\frac{23}{36} - \Sigma$ their 2 probabilities |
### Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X) = \frac{13}{36} + \frac{32}{36} + \frac{12}{36}$ | M1 | Attempt to evaluate $\Sigma xp$ at least 3 values of $x$ in table |
| $= \frac{57}{36}\ \left(\frac{19}{12}\right)\ (1.58)$ | A1 **[2]** | Correct answer |
---4 Coin $A$ is weighted so that the probability of throwing a head is $\frac { 2 } { 3 }$. Coin $B$ is weighted so that the probability of throwing a head is $\frac { 1 } { 4 }$. Coin $A$ is thrown twice and coin $B$ is thrown once.\\
(i) Show that the probability of obtaining exactly 1 head and 2 tails is $\frac { 13 } { 36 }$.\\
(ii) Draw up the probability distribution table for the number of heads obtained.\\
(iii) Find the expectation of the number of heads obtained.
\hfill \mbox{\textit{CAIE S1 2014 Q4 [9]}}