Thornleigh Salesian College, Bolton
Stats Easter School: 2. Probability
Q1.
A group of teenagers bought, in total, 25 items of clothing at two shops, Ace Gear and
Boo Teak.
The following table shows how many tops, jeans and sweaters were bought at each of the
two shops.
Tops
Jeans
Sweaters
Ace Gear
3
7
5
Boo Teak
2
5
3
One item of clothing is chosen at random from these 25 items.
(a)
Find the probability that the chosen item:
(i)
is a top;
(1)
(ii)
was bought from Boo Teak;
(1)
(iii)
is a top and was bought from Boo Teak.
(1)
(b)
State, with a reason, whether the events
‘the chosen item is a top’ and
‘the chosen item was bought from Boo Teak’
are independent.
(2)
(c)
Given that the chosen item is not a top, find the conditional probability that it was
bought from Boo Teak.
(2)
(Total 7 marks)
Page 1
Thornleigh Salesian College, Bolton
Q2.
Gary and his neighbour Larry work at the same place.
On any day when Gary travels to work, he uses one of three options: his car only, a bus
only or both his car and a bus. The probability that he uses his car, either on its own or
with a bus, is 0.6. The probability that he uses both his car and a bus is 0.25.
(a)
Calculate the probability that, on any particular day when Gary travels to work, he:
(i)
does not use his car;
(1)
(ii)
uses his car only;
(2)
(iii)
uses a bus.
(3)
(b)
On any day, the probability that Larry travels to work with Gary is 0.9 when Gary
uses his car only, is 0.7 when Gary uses both his car and a bus, and is 0.3 when
Gary uses a bus only.
(i)
Calculate the probability that, on any particular day when Gary travels to work,
Larry travels with him.
(4)
(ii)
Assuming that option choices are independent from day to day, calculate, to
three decimal places, the probability that, during any particular week (5 days)
when Gary travels to work every day, Larry never travels with him.
(2)
(Total 12 marks)
Page 2
Thornleigh Salesian College, Bolton
Q3.Roger is an active retired lecturer. Each day after breakfast, he decides whether the
weather for that day is going to be fine (F), dull (D) or wet (W). He then decides on only
one of four activities for the day: cycling (C), gardening (G), shopping (S) or relaxing (R).
His decisions from day to day may be assumed to be independent.
The table shows Roger’s probabilities for each combination of weather and activity.
Weather
Fine (F) Dull (D)
Wet
(W)
Cycling (C)
0.30
0.10
0
Gardening (G)
0.25
0.05
0
Shopping (S)
0
0.10
0.05
Relaxing (R)
0
0.05
0.10
Activity
(a)
Find the probability that, on a particular day, Roger decided:
(i)
that it was going to be fine and that he would go cycling;
(ii)
on either gardening or shopping;
(iii)
to go cycling, given that he had decided that it was going to be fine;
(iv)
not to relax, given that he had decided that it was going to be dull;
(v)
that it was going to be fine, given that he did not go cycling.
(9)
(b)
Calculate the probability that, on a particular Saturday and Sunday, Roger decided
that it was going to be fine and decided on the same activity for both days.
(3)
(Total 12 marks)
Page 3
Thornleigh Salesian College, Bolton
Q4.
(a)
The population of a country is suffering from an epidemic of a serious disease.
A hospital doctor is consulted by 250 people suffering from symptoms of this
disease. Of the 250 people, 105 are male, of whom 55 have the disease. Of the
females, 65 have the disease.
A person is selected at random from the 250 people who consulted the hospital
doctor.
The following events are defined:
F is the event “the person is female”;
S is the event “the person has the disease”;
S0 is the event “not S”.
Find:
(i)
P(F);
(ii)
P(F
(iii)
P(S);
(iv)
P(F | S).
S′);
(5)
(b)
The probability that a person selected at random from a large population has the
disease is 0.2.
(i)
Four people are selected at random from a large population. Find the
probability that at least one of these four people has the disease.
(2)
(ii)
Of the people suffering from this disease, 90% give a positive reaction to the
test designed to detect the disease, whereas 15% of the people who are not
suffering from the disease also give a positive reaction to the test.
Calculate the probability that a person who gives a positive reaction to the test
does not have the disease.
(4)
(Total 11 marks)
Page 4
Thornleigh Salesian College, Bolton
M1.
(a)
(i)
P (top) =
B1
1
(ii)
P (from BT) =
B1
1
(iii)
P (top from BT) =
Condone assumption of independence here
B1
1
(b)
, so independent
M1A1
2
(c)
Cond prob =
M1 for
OE
M1A1
2
[7]
M2.
P(C) = 0.6
P(C ∩ B) = 0.25
Page 5
Thornleigh Salesian College, Bolton
{P(C only) = 0.35
(a)
(i)
P(B only) = 0.4}
In (a), ratios (eg 4: 10) are only penalised by
1 mark at first correct answer
P(C′) = 1 – P(C) = 1 – 0.6 = 0.4
CAO; or equivalent
B1
1
(ii)
P(C ∩ B′) = 0.6 – 0.25
Can be implied by correct answer
M1
= 1 – (0.4 + 0.25)
= 0.35
CAO; or equivalent
A1
2
(iii)
P(B) = (i) + p
with p < 0.6
Can be implied by correct answer
M1
= (i) + 0.25
Can be implied by correct answer
A1
= 0.65
CAO; or equivalent
A1
OR
P(B) = 1 – (ii)
Can be implied by correct answer
(M2)
= 0.65
(A1)
Page 6
Thornleigh Salesian College, Bolton
OR
1 = P(C) + P(B) – P(C ∩ B)
Can be implied by correct answer
(M1)
Thus P(B) = 1 – (0.6 – 0.25)
Can be implied by correct answer
(A1)
= 0.65
CAO; or equivalent
(A1)
3
(b)
P(L | GC) = 0.9
P(L | GCB) = 0.7
P(L | GB) = 0.3
(i)
P(G ∩ L)
(a)(ii) × 0.9
(0.315)
Follow through or correct
M1
0.25 × 0.7
(0.175)
M1
[(a)(iii) – 0.25] × 0.3 (0.12)
Follow through or correct
M1
Note: Each pair of multiplied probabilities must be
> 0 to score the corresponding method mark
Ignore any multiplying factors
Ignore any additional terms
0.315 + 0.175 + 0.12 = 0.61
CAO
A1
4
Page 7
Thornleigh Salesian College, Bolton
(ii)
Probability = {1 – (b)(i)}5
Allow 5 × {1 – (b)(i)}5
M1
= 0.395 = 0.009
AWRT (0.00902)
A1
2
[12]
M3.(a)
(i)
P(F & C) = 0.3 or 3 / 10 or 30%
Ratios (eg 3:10) are only penalised by
1 accuracy mark at first correct answer
CAO
(0.3)
B1
(1)
(ii)
P(G or S) = 0.45 or 45 / 100 or 45%
CAO
(0.45)
B1
(1)
(iii)
P(C | F) =
=
M1
30 / 55 or 6 / 11
CAO
(6 / 11)
or
(0.54 to 0.55) or (54% to 55%)
AWFW
(0.54545)
A1
(2)
Page 8
Thornleigh Salesian College, Bolton
(iv)
P(R′ | D) =
Correct numerator
M1
Correct denominator
M1
25 / 30 or 5 / 6
CAO
(5 / 6)
or
(0.83 to 0.834) or (83% to 83.4%)
AWFW
(0.83333)
A1
(3)
(v)
P(F | C′) =
Correct expression
M1
25 / 60 or 5 / 12
CAO
(5 / 12)
or
(0.416 to 0.42) or (41.6% to 42%)
AWRT
(0.41667)
A1
(2,3)
9
(b)
P = [P(F & C)]2 + [P(F & G)]2
Attempt at sum of at least 2 squared
terms; 0 < term < 1; not (a + b)2
May be implied by a correct
expression or a correct answer
M1
0.302 + 0.252 or 0.09 + 0.0625 =
OE
Ignore additional terms or integer multipliers
May be implied by a correct answer
A1
Page 9
Thornleigh Salesian College, Bolton
1525 / 10000 or 305 / 2000 or 61 / 400
CAO
or
(0.1525)
(0.152 to 0.153) or (15.2% to 15.3%)
AWFW
A1
3
[12]
M4.
(a)
(i)
145/250 = 29/50 = 0.58 (or 58%)
B1
(ii)
80/250 = 8/25 = 0.32 (or 32%)
B1
(iii)
120/250 = 12/25 = 0.48 (or 48%)
B1
(iv)
65/120 = 13/24 = 0.542 (or 54.2%)
M1
for using 120
A1
5
(b)
(i)
1 – (0.8)4 = 0.590 = 5904/10000
for 0.84
M1
(or 59.0%)
A1
2
(ii)
P(+ve response) =
(0.2 × 0.9) + (0.8 × 0.15)
M1
Page 10
Thornleigh Salesian College, Bolton
= 0.3
A1
P(not suffering | +ve response) =
(0.8 × 0.15) / 0.3
M1
= 0.12 / 0.3 = 0.4 or 2/5 (or 40%)
A1
4
[11]
Page 11
Thornleigh Salesian College, Bolton
E1.
Once again part (a) was answered very well, and the conditional probability in part (c)
was often found correctly by clear-thinking candidates. Despite this, virtually no candidate
was able to deduce in part (b) that the two events were independent.
E2.
Whilst a not insignificant proportion of candidates scored full or almost full marks,
many candidates found this question difficult; this despite its similarity to a question on a
recent paper. Most candidates answered part (a)(i) correctly but then in parts (a)(ii) and
(iii) often multiplied two probabilities. Similarly, responses in part (b)(i) often involved
multiplying together three probabilities rather than two.
However, where better progress was made, candidates often scored 1 mark for each of
(a)(ii) × 0.9) and (0.25 × 0.7) but not for ((a)(iii) × 0.3) since it required ({(a)(iii) – 0.25} ×
0.3). Such candidates often also scored 1 of the 2 marks available in part (b)(ii) for the use
of (1 – (b)(i))5.
-
E3.This fairly standard probability question posed major difficulties to the weaker students
whereas the stronger students scored high marks often with an apparent minimum of
effort. Almost all students answered part (a)(i) correctly with slightly fewer scoring the
mark in part (a)(ii) through the use of multiplication instead of addition. The use of
conditional probability to answer parts (a)(iii) to (v) was beyond many students and it was
not unusual to see repeated answers of 0.25 or 0.30. Of those students who scored very
few marks in part (a), many were then able to make a sensible and often correct attempt
at part (b). Whilst many students identified the need for (0.302 + 0.252), this was
sometimes spoilt by the introduction of 0.55, 0.55−1 or 2 as a multiplier.
E4.
The main issue is that candidates either like and understand probability concepts and
Page 12
Thornleigh Salesian College, Bolton
achieve full marks on this type of question or, simply leave the question out or make little
effort as they don’t feel confident with the topic. Those who have a go find the conditional
probability hard.
Page 13