| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| 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 distribution construction requiring systematic enumeration of outcomes (8 cases) and basic probability multiplication. Part (ii) involves standard mean and variance calculations from a distribution table. While multi-step, it requires only routine application of independence and probability formulas without novel insight or complex problem-solving. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0) = 0.6 \times 0.25 \times 0.5 = 0.075\) | B1 | 0, 1, 2, 3 seen as top line of a pdf table OR attempting to evaluate \(P(0)\), \(P(1)\), \(P(2)\) and \(P(3)\) |
| \(P(1) = 0.4\times0.25\times0.5 + 0.6\times0.75\times0.5 + 0.6\times0.25\times0.5 = 0.35\) | M1 | Multiply 3 probabilities together from 0.4 or 0.6, 0.25 or 0.75, 0.5 with or without a table |
| \(P(2) = 0.4\times0.75\times0.5 + 0.4\times0.25\times0.5 + 0.6\times0.75\times0.5 = 0.425\) | M1 | Summing 3 probabilities for \(P(1)\) or \(P(2)\) with or without a table |
| \(P(3) = 0.4\times0.75\times0.5 = 0.15\) | B1 | One correct probability seen |
| No of heads | 0 | 1 |
| Prob | \(0.075\ \left(\frac{3}{40}\right)\) | \(0.35\ \left(\frac{7}{20}\right)\) |
| A1 | All correct in a table | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = 0.35 + 2\times0.425 + 3\times0.15 = 1.65\ \left(\frac{33}{20} \text{ oe}\right)\) | M1 | Correct unsimplified expression for the mean using their table, \(\sum p = 1\); can be implied by correct answer |
| \(\text{Var}(X) = 0.35 + 4\times0.425 + 9\times0.15 - 1.65^2\) | M1 | Correct unsimplified expression for the variance using their table and their mean\(^2\) subtracted, \(\sum p = 1\) |
| \(= 0.678\ \left(0.6775\right)\ \left(\frac{271}{400} \text{ oe}\right)\) | A1 | Correct answer |
| Total: 3 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0) = 0.6 \times 0.25 \times 0.5 = 0.075$ | B1 | 0, 1, 2, 3 seen as top line of a pdf table OR attempting to evaluate $P(0)$, $P(1)$, $P(2)$ and $P(3)$ |
| $P(1) = 0.4\times0.25\times0.5 + 0.6\times0.75\times0.5 + 0.6\times0.25\times0.5 = 0.35$ | M1 | Multiply 3 probabilities together from 0.4 or 0.6, 0.25 or 0.75, 0.5 with or without a table |
| $P(2) = 0.4\times0.75\times0.5 + 0.4\times0.25\times0.5 + 0.6\times0.75\times0.5 = 0.425$ | M1 | Summing 3 probabilities for $P(1)$ or $P(2)$ with or without a table |
| $P(3) = 0.4\times0.75\times0.5 = 0.15$ | B1 | One correct probability seen |
| No of heads | 0 | 1 | 2 | 3 |
|-------------|---|---|---|---|
| Prob | $0.075\ \left(\frac{3}{40}\right)$ | $0.35\ \left(\frac{7}{20}\right)$ | $0.425\ \left(\frac{17}{40}\right)$ | $0.15\ \left(\frac{3}{20}\right)$ |
| | A1 | All correct in a table |
| **Total: 5** | | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = 0.35 + 2\times0.425 + 3\times0.15 = 1.65\ \left(\frac{33}{20} \text{ oe}\right)$ | M1 | Correct unsimplified expression for the mean using their table, $\sum p = 1$; can be implied by correct answer |
| $\text{Var}(X) = 0.35 + 4\times0.425 + 9\times0.15 - 1.65^2$ | M1 | Correct unsimplified expression for the variance using their table and their mean$^2$ subtracted, $\sum p = 1$ |
| $= 0.678\ \left(0.6775\right)\ \left(\frac{271}{400} \text{ oe}\right)$ | A1 | Correct answer |
| **Total: 3** | | |5 A game is played with 3 coins, $A , B$ and $C$. Coins $A$ and $B$ are biased so that the probability of obtaining a head is 0.4 for coin $A$ and 0.75 for coin $B$. Coin $C$ is not biased. The 3 coins are thrown once.\\
(i) Draw up the probability distribution table for the number of heads obtained.\\
(ii) Hence calculate the mean and variance of the number of heads obtained.\\
\hfill \mbox{\textit{CAIE S1 2018 Q5 [8]}}