| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | November |
| Marks | 11 |
| 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 S1 question requiring basic probability distribution construction from a given scenario. Part (i) uses binomial distribution with simple calculation. Parts (ii)-(iv) involve counting outcomes from a provided table, calculating probabilities, expectation by formula, and a final probability—all standard textbook exercises with no novel insight required. Slightly easier than average due to the table being given and clear structure. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02b Expectation and variance: discrete random variables |
| 1 | 3 | 5 | 5 | 6 | 6 | ||
| \cline { 2 - 8 } | 1 | 2 | 4 | 6 | 6 | 7 | 7 |
| 3 | 4 | 6 | 8 | 8 | 9 | 9 | |
| First | 5 | 6 | 8 | 10 | 10 | 11 | 11 |
| throw | 5 | 6 | 8 | 10 | 10 | 11 | 11 |
| 6 | 7 | 9 | 11 | 11 | 12 | 12 | |
| 6 | 7 | 9 | 11 | 11 | 12 | 12 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{odd}) = \frac{2}{3}\) or \(0.667\) | B1 | Can be implied if normal approx used with \(\mu = 5.333 (= 8 \times \frac{2}{3})\) |
| \(P(7) = {}_8C_7 \times \left(\frac{2}{3}\right)^7\left(\frac{1}{3}\right) = 0.156\) | M1 | Binomial expression with C in and \(\frac{2}{3}\) and \(\frac{1}{3}\) in powers summing to 8 |
| \(P(8) = \left(\frac{2}{3}\right)^8 = 0.0390\) | M1 | Summing \(P(7) + P(8)\) binomial expressions |
| \(P(7 \text{ or } 8) = 0.195\ (1280/6561)\) | A1 [4] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x\): 2, 4, 6, 7, 8 with \(P(X=x)\): \(\frac{1}{36}, \frac{2}{36}, \frac{5}{36}, \frac{4}{36}, \frac{4}{36}\) | B1 | Values of \(x\) all correct in table of probabilities |
| \(x\): 9, 10, 11, 12 with \(P(X=x)\): \(\frac{4}{36}, \frac{4}{36}, \frac{8}{36}, \frac{4}{36}\) | B2 [3] | All probs correct and not duplicated, \(-1\) ee |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \sum p_i x_i = 2 \times \frac{1}{36} + 4 \times \frac{2}{36} + \ldots\) | M1 | Attempt to find \(\sum p_i x_i\), all \(p < 1\) and no further division of any sort |
| \(= \frac{312}{36}\ \left(\frac{26}{3}\right)\ (8.67)\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > E(X)) = P(X = 9, 10, 11, 12)\) | M1 | Attempt to add their relevant probs |
| \(= \frac{20}{36}\ \left(\frac{5}{9}\right)\ (0.556)\) | A1 [2] | Correct answer |
# Question 7:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{odd}) = \frac{2}{3}$ or $0.667$ | B1 | Can be implied if normal approx used with $\mu = 5.333 (= 8 \times \frac{2}{3})$ |
| $P(7) = {}_8C_7 \times \left(\frac{2}{3}\right)^7\left(\frac{1}{3}\right) = 0.156$ | M1 | Binomial expression with C in and $\frac{2}{3}$ and $\frac{1}{3}$ in powers summing to 8 |
| $P(8) = \left(\frac{2}{3}\right)^8 = 0.0390$ | M1 | Summing $P(7) + P(8)$ binomial expressions |
| $P(7 \text{ or } 8) = 0.195\ (1280/6561)$ | A1 **[4]** | Correct answer |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x$: 2, 4, 6, 7, 8 with $P(X=x)$: $\frac{1}{36}, \frac{2}{36}, \frac{5}{36}, \frac{4}{36}, \frac{4}{36}$ | B1 | Values of $x$ all correct in table of probabilities |
| $x$: 9, 10, 11, 12 with $P(X=x)$: $\frac{4}{36}, \frac{4}{36}, \frac{8}{36}, \frac{4}{36}$ | B2 **[3]** | All probs correct and not duplicated, $-1$ ee |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \sum p_i x_i = 2 \times \frac{1}{36} + 4 \times \frac{2}{36} + \ldots$ | M1 | Attempt to find $\sum p_i x_i$, all $p < 1$ and no further division of any sort |
| $= \frac{312}{36}\ \left(\frac{26}{3}\right)\ (8.67)$ | A1 **[2]** | Correct answer |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > E(X)) = P(X = 9, 10, 11, 12)$ | M1 | Attempt to add their relevant probs |
| $= \frac{20}{36}\ \left(\frac{5}{9}\right)\ (0.556)$ | A1 **[2]** | Correct answer |7 A fair die has one face numbered 1, one face numbered 3, two faces numbered 5 and two faces numbered 6 .\\
(i) Find the probability of obtaining at least 7 odd numbers in 8 throws of the die.
The die is thrown twice. Let $X$ be the sum of the two scores. The following table shows the possible values of $X$.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Second throw}
\begin{tabular}{ l c | c c c c c c }
& & 1 & 3 & 5 & 5 & 6 & 6 \\
\cline { 2 - 8 }
& 1 & 2 & 4 & 6 & 6 & 7 & 7 \\
& 3 & 4 & 6 & 8 & 8 & 9 & 9 \\
First & 5 & 6 & 8 & 10 & 10 & 11 & 11 \\
throw & 5 & 6 & 8 & 10 & 10 & 11 & 11 \\
& 6 & 7 & 9 & 11 & 11 & 12 & 12 \\
& 6 & 7 & 9 & 11 & 11 & 12 & 12 \\
\end{tabular}
\end{center}
\end{table}
(ii) Draw up a table showing the probability distribution of $X$.\\
(iii) Calculate $\mathrm { E } ( X )$.\\
(iv) Find the probability that $X$ is greater than $\mathrm { E } ( X )$.
\hfill \mbox{\textit{CAIE S1 2008 Q7 [11]}}