June 2022 Paper 2 Q13
13 There are 25 students in a class.
- The number of students who study both History and English is 3.
- The number of students who study neither History nor English is 14.
- The number of students who study History but not English is three times the number who study English but not History.
- Show this information on a Venn diagram.
- Determine the probability that a student selected at random studies English. [4]
Two different students from the class are chosen at random.
| Scheme | Marks | AO |
|---|---|---|
| Single Venn diagram drawn showing 3, 14, \(x\) and \(3x\) correctly placed | B1 | 3.1a |
| \(3 + 14 + x + 3x = 25\) oe or \(x = 2\) | M1 | 1.1a |
| Number who study English \(= \text{“}2\text{”} + 3\) or 5 | M1 | 1.1 |
| \(\mathrm{P}(E) = \dfrac{5}{25}\) or \(\dfrac{1}{5}\) or 0.2 | A1 | 1.1 |
| [4] |
Notes
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
M1: May be implied, eg by 2 seen in correct place in diagram
M1: Their \(x + 3\). May be implied by answer
If \(x\) is total English, giving \(x = 5\), use an equivalent scheme.
Alternative (incorrect) method for H \(\leftrightarrow\) E
| Scheme | Marks |
|---|---|
| Diagram | B0 |
| \(3 + 14 + x + 3x = 25\) oe or \(x = 2\) | M1 |
| Number who study English \(= \text{“}6\text{”} + 3\) or 9 | M1 |
| \(\mathrm{P}(E) = \dfrac{9}{25}\) | A1 |
M1: Or implied in diagram, History only = 2, or total History = 5
M1: Their \(3x + 3\). May be implied by answer
A1: If \(x\) is total History, giving \(x = 5\), use an equivalent scheme
| Scheme | Marks | AO |
|---|---|---|
| P(exactly one English) \(= \dfrac{5}{25} \times \dfrac{20}{24} \times 2\) oe | M1 | 1.1 |
| \(= \dfrac{1}{3}\) or 0.333 (3sf) | A1 | 1.1 |
| P(exactly one E and exactly one H) = \(\mathrm{P}(HE' \text{ and } H'E) + \mathrm{P}(EH \text{ and } E'H')\) \(= \left(\dfrac{6}{25} \times \dfrac{2}{24} + \dfrac{3}{25} \times \dfrac{14}{24}\right) \times 2\) oe \(\left(= \dfrac{9}{50} \text{ or } 0.18\right)\) | M2 | 3.1b 2.4 |
| \(\dfrac{\mathrm{P}(\text{exactly one E and exactly one H})}{\mathrm{P}(\text{exactly one E})}\) \(\left(= \dfrac{9}{50} \div \dfrac{1}{3}\right)\) | M1 | 1.1 |
| \(= \dfrac{27}{50}\) or 0.54 (3 sf) cao | A1 | 1.1 |
| [6] |
Notes
M1: Allow omit \(\times 2\). Allow \(\dfrac{5}{25} \times \dfrac{20}{25}\) or 0.16 or 0.32. Allow + ….
A1: NB No ft from (a) in (b)
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 \(\lt 24\))
Allow any of the above + extras for M1
M1: Divide attempted probs of correct events dep \(\geqslant\) M1M1
A1: Careful!! SCs for correct answer by incorrect methods:
“×2” omitted throughout:
\(\dfrac{9}{100} \div \dfrac{1}{6} = \dfrac{27}{50}\): M1A0M1M0M1A1 (Total 4)
Denominator 25×25 instead of 25×24:
\(\dfrac{54}{625} \div \dfrac{4}{25} = \dfrac{27}{50}\): M1A0M1M0M1A1 (Total 4)
Both the above \(\dfrac{27}{625} \div \dfrac{2}{25} = \dfrac{27}{50}\): M1A0M1M0M1A1
Alternative method 1
| Scheme | Marks |
|---|---|
| n(exactly one English) \(= \mathrm{n}(E) \times \mathrm{n}(E')\) | M1 |
| \(= 5 \times 20 = 100\) | A1 |
| n(exactly one E and exactly one H) \(= \mathrm{n}(EH') \times \mathrm{n}(E'H) + \mathrm{n}(EH) \times \mathrm{n}(E'H')\) \(= 2 \times 6 + 3 \times 14\) \((= 54)\) | M1 M1 |
| Attempt \(\dfrac{\mathrm{n}(\text{exactly one E and exactly one H})}{\mathrm{n}(\text{exactly one E})}\) \(\left(= \dfrac{54}{100}\right)\) | M1 |
| \(= \dfrac{27}{50}\) oe or 0.54 (3 sf) cao | A1 |
M1M1: M1 for one of 2×6 or 3×14
or M1 for \(2 \times a + 3 \times b\) (\(a\), \(b\) integers, \(a \lt 23\), \(b \lt 22\))
Alternative method 2
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(H|E) = 3/5\) \(\mathrm{P}(H|E') = 6/20\) \(\mathrm{P}(H'|E) = 2/5\) \(\mathrm{P}(H'|E') = 14/20\) | M1 A1 |
| \(6/20 \times 2/5 + 3/5 \times 14/20\) | M3 |
| \(= 27/50\) | A1 |
M1: M1 for three of these fractions seen
A1: A1 for all four fractions seen
M3: M2 for one of these products \(6/20 \times 2/5\) or \(3/5 \times 14/20\)
or M1 for \(a/20 \times 2/5 + 3/5 \times b/20\)
or M1 for \(6/a \times 2/5 + 3/5 \times 14/b\)
Alternative (incorrect) method for H \(\leftrightarrow\) E
| Scheme | Marks |
|---|---|
| P(exactly one English) \(= \dfrac{9}{25} \times \dfrac{16}{24}\) (×2) | M1 |
| \(= \dfrac{6}{25}\) | A0 |
| P(exactly one E and exactly one H) = \(\mathrm{P}(HE' \text{ and } H'E) + \mathrm{P}(EH \text{ and } E'H')\) Same as main scheme \(\left(= \dfrac{9}{50} \text{ or } 0.18\right)\) | M1 M1 |
| \(\dfrac{\mathrm{P}(\text{exactly one E and exactly one H})}{\mathrm{P}(\text{exactly one E})}\) \(\left(= \dfrac{9}{50} \div \dfrac{6}{25}\right)\) | M1 |
| \(= \dfrac{3}{4}\) or 0.75 (3 sf) | A0 |
M1: Allow without ×2. Allow \(\dfrac{9}{25} \times \dfrac{16}{25}\) or 0.230 or 0.461.
M1: Attempt divide attempted probabilities of correct events dep at least M1M1
A0: SC answer \(\dfrac{3}{4}\), but omit ×2 and/or denominator of 25, M1A0M1M0M1A1