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)
Mark scheme
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
\({}^9C_2\) 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
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
Correct answer only scores full marks (unless from obviously incorrect working)
Answer: \(\dfrac{7}{18}\)
A1ft
(2)
Notes
M1ft: ft diagram, oe
A1ft: ft diagram, oe fraction, decimal or percentage.
NB \(\dfrac{7}{18} = \dfrac{35}{90} = 0.38(88\ldots)\)
Allow ft their tree diagram provided the relevant probabilities are less than 1 in each case
For A1, allow decimals or percentages that round or truncate correctly to at least 2sf. ISW any attempt to convert to other form once correct probability seen
where \(\left[\dfrac{7}{18}\right]\) is their answer to part (b) and must be less than 1
M1ft: ft diagram, for a method to find the probability required. Condone one error in one of the four relevant outcomes or omission of one outcome
Note P(\(RR\)) = P(\(RRR\)) + P(\(RRG\)), so may see \(\dfrac{5}{9} \times \dfrac{7}{10}\left(= \dfrac{7}{18}\right)\) in place of \(\text{``}{\dfrac{7}{22}}\text{''} + \text{``}{\dfrac{7}{99}}\text{''}\) for this mark (similar with P(\(GG\)))
A1ft: ft diagram, correct probability oe fraction, decimal or percentage
NB \(\dfrac{386}{495} = \dfrac{772}{990} = 0.77(979\ldots)\)
Allow ft their tree diagram provided the relevant probabilities are less than 1 in each case
For A1, allow decimals or percentages that round or truncate correctly to at least 2sf. ISW any attempt to convert to other form once correct probability seen
