| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Topic | Modelling and Hypothesis Testing |
| Type | Type I and Type II errors |
| Difficulty | Standard +0.8 This is a multi-stage probability problem requiring careful case analysis, binomial probability calculations, algebraic manipulation to reach a specific form, and expected value computation. While the individual probability concepts are A-level standard, the extended reasoning across multiple stages, the need to simplify to a given expression, and the expected value calculation with conditional sampling make this moderately challenging but still within reach of strong A-level students. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(X = 0) = p^8\) where \(X\) is the number of faulty chips | B1 | State \(p^8\) or \((1-q)^8\) |
| \(P(\text{accept when } X = 1) = 8qp^7 \times p^4\) | M1 | Attempt product of two binomial terms of correct form |
| \(= 8qp^{11}\) | A1 | Correct simplified form seen |
| \(P(\text{accept}) = p^8 + 8p^{11} - 8p^{12}\) | M1 | Use \(q = 1 - p\) to write their expression in terms of \(p\) |
| \(= p^8(1 + 8p^3 - 8p^4)\) | M1 | Sum their two \(np\)'s |
| A1 | [12] | State their \(P(X = 0)\) and \(P(X = 1)\). Obtain given answer |
| (ii) +8 seen in their \(E(X)\) | B1 | Accept \(P(\text{selecting 8}) = 1\). Recognize that (P12) required. |
| \(P(\text{selecting 12}) involving qp^7\) or equiv | M1 | Obtain correct expression. |
| \(= 8qp'\) or \((8p' - 8p^3)\) | A1 | |
| \(E(X) = 8P(8) + 4P(12)\) | M1 | Attempt sum of their two \(np\)'s |
| \(E(X) = 8 + 32qp'\) | A1 | |
| \(E(X) = 9.07\) | A1 | [12] |
(i) $P(X = 0) = p^8$ where $X$ is the number of faulty chips | B1 | State $p^8$ or $(1-q)^8$
$P(\text{accept when } X = 1) = 8qp^7 \times p^4$ | M1 | Attempt product of two binomial terms of correct form
$= 8qp^{11}$ | A1 | Correct simplified form seen
$P(\text{accept}) = p^8 + 8p^{11} - 8p^{12}$ | M1 | Use $q = 1 - p$ to write their expression in terms of $p$
$= p^8(1 + 8p^3 - 8p^4)$ | M1 | Sum their two $np$'s
| A1 | [12] | State their $P(X = 0)$ and $P(X = 1)$. Obtain given answer
(ii) +8 seen in their $E(X)$ | B1 | Accept $P(\text{selecting 8}) = 1$. Recognize that (P12) required.
$P(\text{selecting 12}) involving qp^7$ or equiv | M1 | Obtain correct expression.
$= 8qp'$ or $(8p' - 8p^3)$ | A1 |
$E(X) = 8P(8) + 4P(12)$ | M1 | Attempt sum of their two $np$'s
$E(X) = 8 + 32qp'$ | A1 |
$E(X) = 9.07$ | A1 | [12] | Obtain 9.07 (= 9.06787)In a factory, computer chips are produced in large batches. A quality control procedure is used for each batch which requires a random sample of 8 chips to be tested. If no faulty chip is found, the batch is accepted. If two or more are faulty, the batch is rejected. If one is faulty, a further sample of 4 is selected and the batch is accepted if none of these is faulty. The probability of any chip being faulty is $q$.
\begin{enumerate}[label=(\roman*)]
\item Show that the probability of accepting a batch is $p^8(1 + 8p^3 - 8p^4)$, where $p = 1 - q$. [6]
\item Find the expected number of chips sampled per batch, giving your answer in terms of $p$. Hence show that when $p = 0.75$, the expected number of chips sampled per batch is approximately 9. [6]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q16 [12]}}