Mathematics · May / June 2023 · Paper 61 · Question 4
YMA01
As printed in the paper
The question as text
4. A bag contains a large number of coloured counters. Each counter is labelled A, B or C 30% of the counters are labelled A 45% of the counters are labelled B The rest of the counters are labelled C It is known that 2% of the counters labelled A are red 4% of the counters labelled B are red 6% of the counters labelled C are red One counter is selected at random from the bag.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
4(a)[2]
The answer is a drawing: see the picture of this mark scheme row.
4(b)[2]
0.3 "0.98"
×
= 0.294
(
) (
) (
)
4(c)[2]
0.3 0.02
×
+
"0.45" "0.04"
×
+
"0.25" "0.06"
×
= 0.039
4(d)[3]
P C Red = "0.25" "0.06"
×
0.015
(
)
=
"0.039"
"0.039"
= 0.3846… or 5
13
Notes
(a)
B1 for 0.45, 0.25 and 0.98 Allow fractions
B1 0.04, 0.96 and 0.06, 0.94 Allow fractions
M1 may ft their tree diagram if method shown 0.3 × " their 0.98"
(b)
A1 0.294 oe
(c)
M1 may ft their tree diagram if method shown
A1 0.039 oe
(d)
p
p
M1 allow "their part (c)"
or 0.039
where 0 < p < 1 and p < denominator and their (c) is a
probability or
allow "0.25" "0.06"
×
or 0.015
where 0 < q < 1 and q > numerator
q
q
M1 for "0.25" "0.06"
×
ft their tree diagram and their part(c) if all 3 figures shown in
"0.039"
working. We will condone num > denom
A1 awrt 0.385
NB if correct ft on numerator and denominator leads to “num” > “denom” then max score
is M0M1A0
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.