Higher June 2025 Paper 1R Q19
19 Kannika has 9 counters.
There is a number on each counter.
Kannika puts the 9 counters in a bag.
She takes at random a counter from the bag and does not replace the counter.
She then takes at random a second counter from the bag.
Work out the probability that the sum of the numbers on the two counters is less than 5
(3)
| Scheme | Marks |
|---|---|
\(\dfrac{2}{9} \times \dfrac{1}{8}\left(= \dfrac{2}{72}\right)\) oe or \(\dfrac{1}{9} \times \dfrac{2}{8}\left(= \dfrac{2}{72}\right)\) oe or \(\dfrac{2}{9} \times \dfrac{3}{8}\left(= \dfrac{6}{72}\right)\) oe or \(\dfrac{3}{9} \times \dfrac{2}{8}\left(= \dfrac{6}{72}\right)\) oe or \(\dfrac{2}{9} \times \dfrac{5}{8}\left(= \dfrac{10}{72}\right)\) oe or \(\dfrac{5}{9} \times \dfrac{2}{8}\left(= \dfrac{10}{72}\right)\) oe or \({}^9C_2\) or \(\dfrac{9!}{2!7!}\) or \(\dfrac{9 \times 8}{2}\) or 36 or 1 + 2 + 6 (= 9) | M1 |
\(3 \times \text{``}{\dfrac{2}{72}}\text{''} + 2 \times \text{``}{\dfrac{6}{72}}\text{''}\) oe or \(\text{``}{\dfrac{2}{72}}\text{''} + \text{``}{\dfrac{6}{72}}\text{''} + \text{``}{\dfrac{10}{72}}\text{''}\) oe or \({}^9C_2\) or \(\dfrac{9!}{2!7!}\) or \(\dfrac{9 \times 8}{2}\) or 36 and 1 + 2 + 6 (= 9) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{18}{72}\) | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for finding one correct product
or
for the correct number of total outcomes or for the correct number of outcomes when the sum < 5
NB if using decimals allow 2 decimal places truncated or rounded
M1: for a complete correct method
or
for the correct number of total outcomes and for the correct number of outcomes when the sum < 5
A1: oe eg \(\dfrac{9}{36}\) or 0.25 or 25%
SCB1 for \(\dfrac{21}{81}\) oe eg \(\dfrac{7}{27}\) or 0.259(25…) or 25.9(25…)% truncated or rounded
19 ALT
| Scheme | Marks |
|---|---|
\(\dfrac{2}{9} \times \dfrac{1}{8}\left(= \dfrac{2}{72}\right)\) oe or \(\dfrac{1}{9} \times \dfrac{2}{8}\left(= \dfrac{2}{72}\right)\) or \(\dfrac{1}{9} \times \dfrac{1}{8}\left(= \dfrac{1}{72}\right)\) oe or \(\dfrac{2}{9} \times \dfrac{2}{8}\left(= \dfrac{4}{72}\right)\) or \(\dfrac{3}{9} \times \dfrac{1}{8}\left(= \dfrac{3}{72}\right)\) oe or \(\dfrac{1}{9} \times \dfrac{3}{8}\left(= \dfrac{3}{72}\right)\) oe or \(\dfrac{2}{9} \times \dfrac{3}{8}\left(= \dfrac{6}{72}\right)\) oe or \(\dfrac{3}{9} \times \dfrac{2}{8}\left(= \dfrac{6}{72}\right)\) oe or \(\dfrac{1}{9} \times \dfrac{6}{8}\left(= \dfrac{6}{72}\right)\) oe or \(\dfrac{6}{9} \times \dfrac{1}{8}\left(= \dfrac{6}{72}\right)\) oe or \(\dfrac{3}{9} \times \dfrac{6}{8}\left(= \dfrac{18}{72}\right)\) oe or \(\dfrac{6}{9} \times \dfrac{3}{8}\left(= \dfrac{18}{72}\right)\) oe or \(\dfrac{1}{9} \times \dfrac{8}{8}\left(= \dfrac{8}{72}\right)\) oe or \(\dfrac{8}{9} \times \dfrac{1}{8}\left(= \dfrac{8}{72}\right)\) or \(\dfrac{2}{9} \times \dfrac{8}{8}\left(= \dfrac{16}{72}\right)\) oe or \(\dfrac{8}{9} \times \dfrac{2}{8}\left(= \dfrac{16}{72}\right)\) oe | M1 |
\(1 - \left(2 \times \text{``}{\dfrac{1}{72}}\text{''} + 7 \times \text{``}{\dfrac{2}{72}}\text{''} + 4 \times \text{``}{\dfrac{3}{72}}\text{''} + 2 \times \text{``}{\dfrac{4}{72}}\text{''} + 3 \times \text{``}{\dfrac{6}{72}}\text{''}\right)\) or \(1 - \left(\text{``}{\dfrac{6}{72}}\text{''} + \text{``}{\dfrac{6}{72}}\text{''} + \text{``}{\dfrac{18}{72}}\text{''} + \text{``}{\dfrac{8}{72}}\text{''} + \text{``}{\dfrac{16}{72}}\text{''}\right)\) or \(1 - \dfrac{54}{72}\) oe | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Do not allow \(\dfrac{6}{9} \times \dfrac{3}{8} = \dfrac{18}{72}\) or \(\dfrac{3}{9} \times \dfrac{6}{8} = \dfrac{18}{72}\) as this an incorrect method (M1M0A0) Answer: \(\dfrac{18}{72}\) | A1 |
Notes
M1: for finding one correct product
NB if using decimals allow 2 decimal places truncated or rounded
M1: for a complete correct method
A1: oe eg \(\dfrac{9}{36}\) or 0.25 or 25%
SCB1 for \(\dfrac{21}{81}\) oe eg \(\dfrac{7}{27}\) or 0.259(25…) or 25.9(25…)% truncated or rounded