# Question 1:

## Part (a)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(M=1) = 0.315124\ldots$ awrt $\mathbf{0.315}$ | B1 | |

## Part (a)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(M \geqslant 3) = 1 - P(M \leqslant 2)$ | M1 | M1 for $1 - P(M \leqslant 2)$. Condone writing $1 - P(M < 2)$ if correct answer follows. Just seeing $1 - P(X < 3)$ is not enough unless it leads to correct answer. |
| $= 1 - 0.9885$ awrt $\mathbf{0.0115}$ | A1 | A1 for awrt $0.0115$ (correct answer only 2/2) |

## Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $n \times 0.05 = 3$ (o.e.) | M1 | M1 for writing or using $n \times 0.05$. Can ignore mention of Poisson if correct equation is seen. |
| $n = 60$ | A1 | A1 for 60 only (correct answer only 2/2) |

## Part (c):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $[P(F \geqslant 1) > 0.99 \Rightarrow]\ 1 - P(F=0) > 0.99$ | M1 | 1st M1 for using or writing $1 - P(F=0)$ in a correct inequality or equation **with** $0.99$ |
| $1 - 0.95^n > 0.99$ or $0.95^n < 0.01$ (o.e.) | M1 | 2nd M1 for either of the correct inequalities, allow $\geqslant$ or $\leqslant$ or $=$ (oe). May be implied by 3rd M. **Use of "=" instead of inequality can score 1st two M marks only.** |
| $n \log 0.95 < \log 0.01$ or $n > \dfrac{\log 0.01}{\log 0.95}\ [= 89.78\ldots]$ | M1 | 3rd M1 for solving $0.95^n < 0.01$ (o.e.) (must have an inequality). Must have a correct inequality so $n\log 0.95 = \log 0.01$ is M0A0 even if it leads to 90. For trial and improvement must see both 89 & 90 used. $P(0\|n=89)=0.0104 > 0.01$ and $P(0\|n=90)=0.00988 < 0.01$ then 1st and 2nd M1 implied. |
| $\therefore n = 90$ | A1cso | A1 cso – no sign errors or mistakes |
| **SC** Wrong inequality: $1-P(F=0)<0.99$ leading to $n <$ awrt $89.8$ can score M0M0M1A0 | | |
| **NB** Normal use of normal distribution will score 0/4 | | |

**Total: 9 marks**