| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from given formula |
| Difficulty | Standard +0.3 This is a straightforward discrete probability distribution question requiring calculation of probabilities using basic counting principles (combinations), constructing a probability table, and computing variance using standard formulas. Part (a) is guided (showing a given result), and the methods are all standard S1 techniques with no novel problem-solving required. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=3) = \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5}\) | M1 | \(\frac{m}{7} \times \frac{n}{6} \times \frac{o}{5}\) used throughout; condone use of \(\frac{1}{2}\) |
| \(\frac{6}{35}\) | A1 | AG. The fractions must be identified, e.g. P(NC, NC, C), may be seen in a tree diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Table with \(x\) values 1, 2, 3, 4, 5 and \(p\) values \(\frac{15}{35}\), \(\frac{10}{35}\), \(\frac{6}{35}\), \(\frac{3}{35}\), \(\frac{1}{35}\) | B1 | Table with \(x\) values and at least one probability. Condone any additional \(x\) values if probability stated as 0 |
| B1 | One correct probability other than \(X = 3\) linked to the correct outcome | |
| B1 | Two further correct probabilities other than \(X = 3\) seen linked to the correct outcome | |
| B1FT | All probabilities correct, or at least 4 probabilities summing to 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 1\times\frac{15}{35} + 2\times\frac{10}{35} + 3\times\frac{6}{35} + 4\times\frac{3}{35} + 5\times\frac{1}{35}\) | M1 | At least 4 correct terms FT *their* values in (a) with probabilities summing to 1. May be implied by use in Variance, accept unsimplified expression |
| \(E(X) = \frac{15+20+18+12+5}{35} = \frac{70}{35} = 2\) | ||
| \(\text{Var}(X) = \left[\frac{1^2\times15 + 2^2\times10 + 3^2\times6 + 4^2\times3 + 5^2\times1}{35} - 2^2\right]\) | M1 | Appropriate variance formula using *their* \((E(X))^2\). FT *their* table, accept probabilities not summing to 1 |
| \(\frac{15+40+54+48+25}{35} - 2^2\) | ||
| \(= \frac{182}{35} - 4 = \frac{6}{5}\) | A1 | N.B. If method FT for M marks from *their* incorrect (b), expressions for \(E(X)\) and \(\text{Var}(X)\) must be seen unsimplified with all probabilities \(< 1\) |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=3) = \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5}$ | M1 | $\frac{m}{7} \times \frac{n}{6} \times \frac{o}{5}$ used throughout; condone use of $\frac{1}{2}$ |
| $\frac{6}{35}$ | A1 | AG. The fractions must be identified, e.g. P(NC, NC, C), may be seen in a tree diagram |
**Total: 2 marks**
---
## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Table with $x$ values 1, 2, 3, 4, 5 and $p$ values $\frac{15}{35}$, $\frac{10}{35}$, $\frac{6}{35}$, $\frac{3}{35}$, $\frac{1}{35}$ | B1 | Table with $x$ values and at least one probability. Condone any additional $x$ values if probability stated as 0 |
| | B1 | One correct probability other than $X = 3$ linked to the correct outcome |
| | B1 | Two further correct probabilities other than $X = 3$ seen linked to the correct outcome |
| | B1FT | All probabilities correct, or at least 4 probabilities summing to 1 |
**Total: 4 marks**
---
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 1\times\frac{15}{35} + 2\times\frac{10}{35} + 3\times\frac{6}{35} + 4\times\frac{3}{35} + 5\times\frac{1}{35}$ | M1 | At least 4 correct terms FT *their* values in (a) with probabilities summing to 1. May be implied by use in Variance, accept unsimplified expression |
| $E(X) = \frac{15+20+18+12+5}{35} = \frac{70}{35} = 2$ | | |
| $\text{Var}(X) = \left[\frac{1^2\times15 + 2^2\times10 + 3^2\times6 + 4^2\times3 + 5^2\times1}{35} - 2^2\right]$ | M1 | Appropriate variance formula using *their* $(E(X))^2$. FT *their* table, accept probabilities not summing to 1 |
| $\frac{15+40+54+48+25}{35} - 2^2$ | | |
| $= \frac{182}{35} - 4 = \frac{6}{5}$ | A1 | **N.B.** If method FT for M marks from *their* incorrect (b), expressions for $E(X)$ and $\text{Var}(X)$ must be seen unsimplified with all probabilities $< 1$ |
**Total: 3 marks**7 Sharma knows that she has 3 tins of carrots, 2 tins of peas and 2 tins of sweetcorn in her cupboard. All the tins are the same shape and size, but the labels have all been removed, so Sharma does not know what each tin contains.
Sharma wants carrots for her meal, and she starts opening the tins one at a time, chosen randomly, until she opens a tin of carrots. The random variable $X$ is the number of tins that she needs to open.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 3 ) = \frac { 6 } { 35 }$.
\item Draw up the probability distribution table for $X$.
\item Find $\operatorname { Var } ( X )$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q7 [9]}}