# Question 13:

## Part (a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Single Venn diagram drawn showing $3$, $14$, $x$ and $3x$ correctly placed | B1 | or showing $3$, $14$, $2$ and $6$ correctly placed; Allow omission of rectangle, so long as 14 seen outside; Allow probabilities in the diagram |
| $3 + 14 + x + 3x = 25$ oe or $x = 2$ | M1 | May be implied, eg by 2 seen in correct place in diagram |
| Number who study English $= "2" + 3$ or $5$ | M1 | Their $x + 3$. May be implied by answer |
| $P(E) = \frac{5}{25}$ or $\frac{1}{5}$ or $0.2$ | A1 | If $x$ is total English, giving $x = 5$, use an equivalent scheme |

## Part (a) ctd — Alternative (incorrect) method for $H \leftrightarrow E$:

| Answer | Mark | Guidance |
|--------|------|----------|
| Diagram | B0 | |
| $3 + 14 + x + 3x = 25$ oe or $x = 2$ | M1 | Or implied in diagram, History only $= 2$, or total History $= 5$ |
| Number who study English $= "6" + 3$ or $9$ | M1 | Their $3x + 3$. May be implied by answer |
| $P(E) = \frac{9}{25}$ | A1 | If $x$ is total History, giving $x = 5$, use an equivalent scheme |

## Part (b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{exactly one English}) = \frac{5}{25} \times \frac{20}{24} \times 2$ oe $= \frac{1}{3}$ or $0.333$ (3sf) | M1, A1 | Allow omit $\times 2$. Allow $\frac{5}{25} \times \frac{20}{25}$ or $0.16$ or $0.32$. Allow $+\ldots$; NB No ft from (a) in (b) |
| $P(\text{exactly one E and exactly one H}) = P(HE' \text{ and } H'E) + P(EH \text{ and } E'H')$ $= \left(\frac{6}{25} \times \frac{2}{24} + \frac{3}{25} \times \frac{14}{24}\right) \times 2$ oe | M2 | M1 for one of $\frac{6}{25} \times \frac{2}{24}$ or $\frac{3}{25} \times \frac{14}{24}$ oe; OR $\frac{6}{25} \times \frac{2}{25} + \frac{3}{25} \times \frac{14}{25}$ (both terms); OR $\frac{6}{25} \times \frac{a}{24} + \frac{3}{25} \times \frac{b}{24}$ or $\frac{2}{25} \times \frac{a}{24} + \frac{14}{25} \times \frac{b}{24}$ ($a,b$ integer $< 24$); Allow any of the above $+$ extras for M1 |
| $\left(= \frac{9}{50} \text{ or } 0.18\right)$ | | |
| $\frac{P(\text{exactly one E and exactly one H})}{P(\text{exactly one E})} = \frac{9}{50} \div \frac{1}{3}$ | M1 | Divide attempted probs of correct events dep $\geq$ M1M1 |
| $= \frac{27}{50}$ or $0.54$ (3 sf) cao | A1 | Careful!! SCs for correct answer by incorrect methods: "$\times 2$" omitted throughout: $\frac{9}{100} \div \frac{1}{6} = \frac{27}{50}$: M1A0M1M0M1A1 (Total 4); Denominator $25\times25$ instead of $25\times24$: $\frac{54}{625} \div \frac{4}{25} = \frac{27}{50}$: M1A0M1M0M1A1; Both the above $\frac{27}{625} \div \frac{2}{25} = \frac{27}{50}$: M1A0M1M0M1A1 |

## Part (b) ctd — Alternative method 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| $n(\text{exactly one English}) = n(E) \times n(E') = 5 \times 20 = 100$ | M1, A1 | |
| $n(\text{exactly one E and exactly one H}) = n(EH') \times n(E'H) + n(EH) \times n(E'H')$ $= 2\times6 + 3\times14 \quad (= 54)$ | M1, M1 | M1 for one of $2\times6$ or $3\times14$; or M1 for $2\times a + 3\times b$ ($a, b$ integers, $a < 23$, $b < 22$) |
| Attempt $\frac{n(\text{exactly one E and exactly one H})}{n(\text{exactly one E})}$ | M1 | |
| $\left(= \frac{54}{100}\right) = \frac{27}{50}$ oe or $0.54$ (3 sf) cao | A1 | |

## Part (b) ctd — Alternative method 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $P(H/E) = 3/5 \quad P(H/E') = 6/20$ | M1 | M1 for three of these fractions seen |
| $P(H'/E) = 2/5 \quad P(H'/E') = 14/20$ | A1 | A1 for all four fractions seen |
| $6/20\times2/5 + 3/5\times14/20$ | M3 | M2 for one of these products $6/20\times2/5$ or $3/5\times14/20$; or M1 for $a/20\times2/5 + 3/5\times b/20$; or M1 for $6/a\times2/5 + 3/5\times14/b$ |
| $= 27/50$ | A1 | |

## Question 13(b): Alternative (incorrect) method for H↔E

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{exactly one English}) = \frac{9}{25} \times \frac{16}{24} \times 2$ | M1 | Allow without $\times 2$. Allow $\frac{9}{25} \times \frac{16}{25}$ or 0.230 or 0.461 |
| $= \frac{6}{25}$ | A0 | |
| $P(\text{exactly one E and exactly one H}) = P(HE' \text{ and } H'E) + P(EH \text{ and } E'H')$ | M1 | |
| Same as main scheme | M1 | |
| $= \frac{9}{50}$ or $0.18$ | | |
| $\dfrac{P(\text{exactly one E and exactly one H})}{P(\text{exactly one E})}$ | M1 | Attempt divide attempted probabilities of correct events; dep at least M1M1 |
| $= \frac{9}{50} \div \frac{6}{25}$ | | |
| $= \frac{3}{4}$ or $0.75$ (3 sf) | A0 | SC answer $\frac{3}{4}$, but omit $\times 2$ and/or denominator of 25: M1A0M1M0M1A1 |

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