| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 10 |
| Topic | Combinations & Selection |
| Type | Conditional probability in selection |
| Difficulty | Standard +0.3 This is a multi-part question covering basic probability distributions and straightforward combinatorics. Part (i)-(iii) involve simple algebra with probability axioms and standard expectation formulas. Parts (iv)-(v) require conditional probability but with clear setups. The combinatorics sections use standard techniques (complementary counting, arrangements with constraints, selection with restrictions). While lengthy, each individual part is routine and requires no novel insight—slightly above average difficulty only due to the number of parts and minor conditional probability reasoning. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 1 | 3 | 5 | 7 | 9 |
| \(\mathrm { P } ( X ) = x\) | 0.3 | \(p\) | 0.2 | \(2 p\) | 0.2 |
(i) $p = 0.1$ [B1] **1 mark**
(ii) $E(X) = 1\times0.3 + 3\times0.1 + 5\times0.2 + 7\times0.2 + 9\times0.2 = 4.8$ [B1]
$E(X^2) = 1\times0.3 + 9\times0.1 + 25\times0.2 + 49\times0.2 + 81\times0.2 = 32.2$ [B1] **2 marks**
(iii) Use of the formula $\text{Var}(X) = E(X^2) - (E(X))^2$ [M1]
Obtain $32.2 - 4.8^2 = 9.16$ [A1] **2 marks**
(iv) Implication of the applicability of the arithmetic of Mutually Exclusive and Independent events [M1]
Probability of both prime is $(0.1 + 0.2 + 0.2) \times (0.2 + 0.2 + 0.2) = 0.3$ [A1] **2 marks**
(v) Clear definitions for the following events:
$A =$ 'Both generate a power of 3, namely 1, 3 or 9'
$B =$ 'The first generates 1, 3 or 5'
$A \cap B =$ 'The first generates 1 or 3 and the second generates 1, 3 or 9' [B1]
Use of the conditional probability formula $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ [M1]
Obtain $\frac{0.4 \times 0.6}{0.6} = 0.4$ [A1] **3 marks**12 A faulty random number generator generates odd digits according to the probability distribution for the random variable $X$ given in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 3 & 5 & 7 & 9 \\
\hline
$\mathrm { P } ( X ) = x$ & 0.3 & $p$ & 0.2 & $2 p$ & 0.2 \\
\hline
\end{tabular}
\end{center}
(i) Find $p$.\\
(ii) Find $\mathrm { E } ( X )$ and $\mathrm { E } \left( X ^ { 2 } \right)$.\\
(iii) Deduce the value of $\operatorname { Var } ( X )$.
A second random number generator generates odd digits each with equal probability. Both random generators are operated once.\\
(iv) Find the probability that both generate a prime number.\\
(v) Given that the first generates 1, 3 or 5, find the probability that both generate a power of 3 .
1315 pupils, including two sisters, are placed in random order in a line.\\
(i) What is the probability that the sisters are not next to each other?\\
(ii) How many arrangements are there with 9 pupils between the sisters?
A team of 5 is chosen from the 15 pupils.\\
(iii) How many ways are there of choosing the team if no more than one of the sisters can be in the team?
Having chosen the first team, a second team of 5 pupils is chosen from the remaining 10 pupils.\\
(iv) How many ways are there of choosing the teams if each sister is in one or other of the teams?
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q12 [10]}}