| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Topic | Binomial Distribution |
| Type | At least one success |
| Difficulty | Moderate -0.3 This is a straightforward binomial probability question requiring standard formula application. Part (i) involves direct use of binomial probability with n=10, p=0.2 for cumulative and individual probabilities. Part (ii) requires solving 1-(0.8)^n ≥ 0.95 using logarithms. All techniques are routine A-level content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim \mathrm{B}(10, 0.2)\), \(\mathrm{P}(X < 3) = 0.8791\) | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathrm{P}(X = 3) = 0.8791 - 0.6778 = 0.2013\) | M1, A1 | Use of tables for \(\mathrm{P}(X \leq 3) - \mathrm{P}(X \leq 2)\) or formula for \(\mathrm{P}(X = 3)\). c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathrm{P}(X > 1) = 1 - 0.1074 = 0.8926\) | M1, A1 | Attempt \(1 - \mathrm{P}(X \leq 0)\) using tables or formula for \(\mathrm{P}(X = 0)\). c.a.o |
| Answer | Marks |
|---|---|
| \(X \sim \mathrm{B}(n, 0.2)\). Require \(\mathrm{P}(X > 1) > 0.95\), i.e. require \(n\) s.t. \(\mathrm{P}(X > 1) = 1 - 0.8^n > 0.95\) or s.t. \(0.8^n < 0.05\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| If \(n = 13\), \(\mathrm{P}(X > 1) = 1 - 0.0549\ldots = 0.9450\ldots < 0.95\) | A1 | Either trial: \(n = 13\) or 14. The outcome of the trial must be shown explicitly. |
| If \(n = 14\), \(\mathrm{P}(X > 1) = 1 - 0.0439\ldots = 0.9570\ldots > 0.95\) | A1 | Both trials shown explicitly and the conclusion stated. |
| \(\therefore n = 14\) | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(n \log 0.8 \leq \log 0.05\) | A1 | Takes logs, any base, and make \(n\) the subject. |
| \(\therefore n = \frac{\log 0.05}{-1.3010} = \frac{-2.9957}{-0.2231} = 13.425\) | A1 | Correct value and conclusion. |
### (i)
#### (a)
$X \sim \mathrm{B}(10, 0.2)$, $\mathrm{P}(X < 3) = 0.8791$ | B1 | [1]
#### (b)
$\mathrm{P}(X = 3) = 0.8791 - 0.6778 = 0.2013$ | M1, A1 | Use of tables for $\mathrm{P}(X \leq 3) - \mathrm{P}(X \leq 2)$ or formula for $\mathrm{P}(X = 3)$. c.a.o. | [2]
#### (c)
$\mathrm{P}(X > 1) = 1 - 0.1074 = 0.8926$ | M1, A1 | Attempt $1 - \mathrm{P}(X \leq 0)$ using tables or formula for $\mathrm{P}(X = 0)$. c.a.o | [2]
### (ii)
$X \sim \mathrm{B}(n, 0.2)$. Require $\mathrm{P}(X > 1) > 0.95$, i.e. require $n$ s.t. $\mathrm{P}(X > 1) = 1 - 0.8^n > 0.95$ or s.t. $0.8^n < 0.05$ | M1 |
**ALT version 1, by trial and error:**
If $n = 13$, $\mathrm{P}(X > 1) = 1 - 0.0549\ldots = 0.9450\ldots < 0.95$ | A1 | Either trial: $n = 13$ or 14. The outcome of the trial must be shown explicitly.
If $n = 14$, $\mathrm{P}(X > 1) = 1 - 0.0439\ldots = 0.9570\ldots > 0.95$ | A1 | Both trials shown explicitly and the conclusion stated.
$\therefore n = 14$ | [3]
**ALT version 2, by logs:**
$n \log 0.8 \leq \log 0.05$ | A1 | Takes logs, any base, and make $n$ the subject.
$\therefore n = \frac{\log 0.05}{-1.3010} = \frac{-2.9957}{-0.2231} = 13.425$ | A1 | Correct value and conclusion. |
---A certain type of sweet is made in a variety of colours. $20\%$ of the sweets made are blue. Sweets of the various colours are thoroughly mixed before being put into packets.
\begin{enumerate}[label=(\roman*)]
\item In a packet that contains 10 sweets, find the probability that the packet contains
\begin{enumerate}[label=(\alph*)]
\item at most 3 blue sweets, [1]
\item exactly 3 blue sweets, [2]
\item at least 1 blue sweet. [2]
\end{enumerate}
\item What is the smallest number of sweets that a packet should contain in order to be at least $95\%$ certain of having at least 1 blue sweet? [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q4 [8]}}