| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring basic construction of a table, calculation of expectation and variance using standard formulas, and application of binomial distribution. The only mild challenge is correctly interpreting 'coins gained' as net profit (accounting for the £1 cost), but the multi-part structure guides students through each step methodically. Slightly easier than average due to its routine nature and clear scaffolding. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
**(i)** Table shows $(-1, 0.7)$, $(0, 0.25)$ and $(9, 0.05)$ — B1, B1 [2]
**(ii)** Use $E(X)$ formula — M1
Obtain $-0.25$ AG — A1
Use $E(X^2)$ formula — M1
Obtain 4.69 or $\left(\frac{75}{16}\right)$ — A1 [4]
**(iii)** Use $10 + 10E(X)$ — M1
Obtain $10 + 10(-0.25) = 7.5$ — A1 [2]
**(iv)** P(Must win at least one game) — M1
State $(0.25)^{10}$ — B1
Obtain $1 - (0.95)^{10} + (0.25)^{10} = 0.401$ — A1 [3]
OR $1 - \sum_{r=0}^{r=9}\left(^{10}C_r\, 0.7^{10-r} 0.25^r\right)$ — M1 *(Summation of attempt at relevant terms)*
$= 1 - 0.59873\ldots$ — A1 *(All terms correct)*
$= 0.401(26\ldots)$ — A1
**Total: [11]**
*SC For (0, 1, 10) allow max B1B0. $(0.25)^{10}$ must be seen in the final calculation though it does not affect the value of 0.401. Do not ISW if cand rounds answer (to 7 or 8).*6 James plays an arcade game. Each time he plays, he puts a $\pounds 1$ coin in the slot to start the game. The possible outcomes of each game are as follows:
James loses the game with a probability of 0.7 and the machine pays out nothing, James draws the game with a probability of 0.25 and the machine pays out a $\pounds 1$ coin, James wins the game with a probability of 0.05 and the machine pays out ten $\pounds 1$ coins.
The outcomes can be modelled by a random variable $X$ representing the number of $\pounds 1$ coins gained at the end of a game.\\
(i) Construct a probability distribution table for $X$.\\
(ii) Show that $\mathrm { E } ( X ) = - 0.25$ and find $\operatorname { Var } ( X )$.
James starts off with $10 \pounds 1$ coins and decides to play exactly 10 games.\\
(iii) Find the expected number of $\pounds 1$ coins that James will have at the end of his 10 games.\\
(iv) Find the probability that after his 10 games James will have at least $10 \pounds 1$ coins left.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2012 Q6 [11]}}