# Question 5:

## Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{at least 3 whites}) = (1-0.07)^3$ or $1-0.07-0.93\times0.07-0.93^2\times0.07$ | M1 | 1.1b - A correct method to find $P(X \geq 3)$ |
| $= 0.8043...$ awrt 0.804 | A1 | 1.1b |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{2nd red on 9th draw}) = \dbinom{8}{1}0.93^7\times0.07^2$ | M1 | 3.3 - For selecting appropriate model: negative binomial or binomial with extra trial |
| $= 0.02358...$ awrt 0.0236 | A1 | 1.1b |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{n}{p} = 4400$ and $\dfrac{n(1-p)}{p^2} = 660^2$ | M1, A1 | 3.1b, 1.1b - Forming equation for mean and variance. At least one correct. Both equations correct. Allow M1A1 if both equations correct with same number substituted for $n$ |
| $1-p = 99p$ oe | M1 | 1.1b - Solving 2 equations leading to $1-p=99p$. Allow $p-p^2=99p^2$ ft their 4400 and 660. Allow $1-p=0.15p$ |
| $p = 0.01$ | A1 | 1.1b |

## Part (d):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: p = 0.07$, $H_1: p < 0.07$ | B1 | 2.5 - Both hypotheses correct using correct notation, allow e.g. $p > 0.93$ |
| $J \sim \text{Geo}(0.07)$ | M1 | 3.3 - Realising need to use $\text{Geo}(0.07)$ ft their hypotheses |
| $P(J \geq c) < 0.1 \Rightarrow (1-0.07)^{c-1} < 0.1$ | M1 | 3.4 - Using model to find $P(J \geq c)$. Condone $(1-0.07)^c < 0.1$ ft their $0.07 \neq 0.93$. ALT $P(J\geq32)=0.1[054...]$ or $P(J\geq33)=0.09[8...]$ implied by correct CR |
| $c - 1 > \dfrac{\log 0.1}{\log 0.93}$ | M1 | 1.1b - Valid method to solve inequality or $P(J\geq32)=0.1[054]$ and $P(J\geq33)=0.09[81]$ implied by correct CR |
| $c > 32.72...$ $\therefore$ CR: $J \geq 33$ | A1 | 1.1b - Correct CR (any letter). A0 if given as probability statement. Must be integer |

## Part (e):

| Answer/Working | Marks | Guidance |
|---|---|---|
| 34 is in the critical region | M1 | 1.1b - Comparing 34 with their CR e.g. $34>33$, $34\geq33$ or $P(J\geq34)=0.09[12]$ |
| There is evidence to suggest that Jerry's bag contains a smaller proportion of red counters than Asha's bag | A1 | 2.2b - Fully correct conclusion in context. Allow Jerry's belief is true. Allow probability for proportion |

## Part (f):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Power of test $= P(J \geq 33 \mid p = 0.011)$ | M1 | 2.1 - Realising need to find $P$(their CR in (d)). Allow $1-P(J\leq32)$ |
| $= (1-0.011)^{32}$ oe | M1 | 1.1b - Correct method. Allow $1-0.2981...$. May be implied by $0.7019...$. If CR incorrect $(1-0.011)^{\text{"CR"}-1}$ or $1-\{1-(1-0.011)^{\text{"CR"}-1}\}$ must be seen |
| $= 0.7019...*$ | A1* | 1.1b - Only award if both method marks awarded |

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