## (i)

**(A)** $P(\text{both rest of UK}) = 0.20 \times 0.20 = 0.04$ | M1, A1cao | M1 for multiplying; A1cao | 2 marks

**(B)** **Either:** All 5 case

$P(\text{at least one England}) = (0.79 \times 0.20) + (0.79 \times 0.01) + (0.20 \times 0.79) + (0.01 \times 0.79) + (0.79 \times 0.79)$

$= 0.158 + 0.0079 + 0.158 + 0.0079 + 0.6241 = 0.9559$ | M1 | M1 for any correct term (3case or 5case); M1 for correct sum of all 3 (or all 5) with no extras; A1cao (condone 0.96 www)

**Or**

$P(\text{at least one England}) = 1 - P(\text{neither England})$
$= 1 - (0.21 \times 0.21) = 1 - 0.0441 = 0.9559$ | M1, M1dep | Or M1 for $0.21 \times 0.21$ or for (*) fully enumerated or 0.0441 seen; M1dep for 1 − (1st part); A1cao

**Or:** All 3 case

$P(\text{at least one England}) = 0.79 \times 0.21 + 0.21 \times 0.79 + 0.79^2$
$= 0.1659 + 0.1659 + 0.6241 = 0.9559$ | M1 | See above for 3 case | 3 marks

**(C)** **Either:**

$0.79 \times 0.79 + 0.79 \times 0.2 + 0.2 \times 0.79 + 0.2 \times 0.2 = 0.9801$ | M1 | M1 for sight of all 4 correct terms summed; A1cao (condone 0.98 www)

**Or**

$0.99 \times 0.99 = 0.9801$ | M1, A1cao | Or: M1 for $0.99 \times 0.99$; A1cao

**Or**

$1 - (0.79 \times 0.01 + 0.2 \times 0.01 + 0.01 \times 0.79 + 0.01 \times 0.02 + 0.01^2) = 1 - 0.0199 = 0.9801$ | M1 | Or: M1 for everything 1 − {.....}; A1cao | 2 marks total

## (ii)

$P(\text{both the rest of the UK} \mid \text{neither overseas}) = \frac{P(\text{the rest of the UK and neither overseas})}{P(\text{neither overseas})} = \frac{0.04}{0.9801} = 0.0408$ | M1, M1, A1 | M1 for numerator of 0.04 or 'their answer to (i)(A)'; M1 for denominator of 0.9801 or 'their answer to (i)(C)'; A1 FT ($0 < p < 1$) 0.041 at least | 3 marks

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## (iii)

**(A)** Probability $= 1 - 0.79^5 = 1 - 0.3077 = 0.6923$ (accept awrt 0.69) | M1, M1, A1 | M1 for $0.79^5$ or 0.3077...; M1 for 1 − 0.79^5 dep; A1cao

**See additional notes for alternative solution**

**(B)** $1 - 0.79^n > 0.9$ | M1 | M1 for equation/inequality in n (accept either statement opposite)

**Either:**

$1 - 0.79^n > 0.9$ or $0.79^n < 0.1$ (condone = and ≥ throughout) but not reverse inequality | M1(indep), A1 | M1(indep) for process of using logs i.e. $\frac{\log a}{\log b}$

$n > \frac{\log 0.1}{\log 0.79}$, so $n > 9.768...$ | A1cao | A1cao

Minimum $n = 10$ Accept $n \geq 10$ | 3 marks

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**Or** (using trial and improvement):

Trial with $0.79^9$ or $0.79^{10}$

$1 - 0.79^9 = 0.8801 (< 0.9)$ or $0.79^9 = 0.1198 (> 0.1)$ | M1(indep) | M1(indep) for sight of 0.8801 or 0.1198

$1 - 0.79^{10} = 0.9053 (> 0.9)$ or $0.79^{10} = 0.09468 (< 0.1)$ | M1, A1 | M1 for sight of 0.9053 or 0.09468; A1 dep on both M's cao

Minimum $n = 10$ Accept $n \geq 10$ | 3 marks total

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**NOTE:** $n = 10$ unsupported scores SC1 only | | 16 marks total

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