Answers: Apply probability concepts in solving problems (91585)
One
(a) (i)
(b) (i)
Expected Coverage
P(from Australia) = 0.48
P(on holiday/from Australia) = 0.63
P(both islands/on holiday/from Australia) =
0.27
(this may be drawn on a probability tree)
P(visitor from AUS on business)
= 0.48 x 0.37
= 0.1776
P(both islands ∩ on holiday ∩ from Australia)
= 0.48 × 0.63 × 0.27
= 0.0816 (3 s.f.)
P(both from Australia on business) = 0.17762 = 0.0315 (3 s.f.)
(b) (ii)
The assumption is that of independence, i.e. because of the large number of
people in the study, selecting one person does not change the proportion of
0.1776 significantly.
Prepared by TEAM Solutions University of Auckland
Achievement
Merit
Candidate calculates
probability correctly.
Candidate calculates
probability correctly.
Candidate calculates
probability correctly
Candidate calculates
probability correctly
AND
AND
states assumption of
independence for both
people being Australians
on a business trip
explains why
assumption of
independence is
reasonable.
Excellence
Student constructs a two-way table (or uses equivalent other diagrams or
probability statements):
From
Australia
Not from
Australia
Visits both
islands
Visits one
island only
Total
14
58
72
10
68
78
24
126
150
Candidate calculates
probability correctly
AND
states the events are not
mutually exclusive for
part (i).
Candidate determines
the count / probability of
another combined event
not given in the question
(partial completion of
Venn diagram) for part
(ii).
Candidate calculates
the probability
correctly for part (ii).
(c) (i)
Total
P(from Australia ∩ both islands) = 14 ÷ 150 = 0.0933 (3 s.f.) or fractional
form.
As this is not zero, the two events are not mutually exclusive.
(c) (ii)
P(on holiday ∩ both islands) = 17 ÷ 150 = 0.113 (3 s.f.) or fractional form
Two
Expected Coverage
P(purchase) = 0.113 + 0.214 = 0.327
P(tour group) = 0.214 + 0.452 = 0.666
(a) (i)
P(purchase ∩ tour group) = 0.214
P(purchase) × P(tour group) = 0.327 × 0.666 = 0.218 (3 s.f.) ≠ 0.214
Therefore the events are not independent.
Achievement
Independence rule used
with correct probabilities
to determine events are
not independent.
Accept alternatives, eg
P(B/A) ≠ P(B).
Merit
Excellence
P(purchase/independent) = 0.113 ÷ 0.334 = 0.338 (3 s.f.)
P(purchase/tour group) = 0.214 ÷ 0.666 = 0.321 (3 s.f.)
One conditional
probability correctly
calculated.
Both conditional
probabilities calculated
and compared to reach
conclusion.
(a) (ii)
Accept alternative
reasoning, eg relative
ratios.
Yes, an independent traveller is slightly more likely to make a purchase than
a traveller in a tour group.
(b) (i)
P(member and no purchase) = 5/50 = 0.1
P(have flu) = 9 ÷ 68400
P(identified through screening and have flu) = 13 ÷ 10000 × 0.1 = 0.00013
(c)
P(not identified through screening and have flu) = 9 ÷ 68400 – 0.00013
Proportion of people
identified through
screening and who have
the flu correctly
calculated.
Proportion of people not
identified through
screening and who have
the flu correctly
calculated.
Approximate risk of not
being diagnosed with
the flu through the
screening process
correctly calculated.
Achievement
Merit
Excellence
Risk of not being identified while having flu
= P(not identified through screening/have flu)
= (9 ÷ 68400 – 0.00013) ÷ (9 ÷ 68400)
= 0.012
The risk of not being identified in the screening process is 1.2%.
Three
(a)(i)
and
(a)(ii)
Expected Coverage
Model probability:
P(two attempts) = 1/6 x 1/8 = 1/48
P(more than two attempts) = 1 –1/48= 47/48 or 0.9792
Probability from graph:
P(two attempts) = 12/500
P(more than two attempts) = 1 – 12/500 = 0.977
This is similar to the observed result of 97.7%.
Either model or
experimental probability
calculated.
Both model and
experimental
probabilities calculated
with comment on
similarity.
P(five pictures) = P(4 for first geyser and 1 for second geyser)
+ P(3 for first geyser and 2 for second geyser)
+ P(2 for first geyser and 3 for second geyser)
+ P(1 for first geyser and 4 for second geyser)
(b)
P(4 for first geyser and 1 for second geyser)
= (1 –1/6)3 × 1/6 × 1/8 = 0.01206 (4 s.f.)
P(3 for first geyser and 2 for second geyser)
= (1 –1/6)2 × 1/6 × (1 –1/8) × 1/8= 0.01266(4 s.f.)
P(2 for first geyser and 3 for second geyser)
= (1 – 1/6) × 1/6× (1 –1/8) 2 × 1/8= 0.01329(4 s.f.)
P(1 for first geyser and 4 for second geyser)
= 1/6× (1 –1/8) 3 × 1/8= 0.1396(4 s.f.)
P(five pictures) = 0.0520(3 s.f.)
One possibility for
taking five attempts
identified
All possibilities for
taking five attempts
identified
AND
AND
one probability
calculated.
at least one probability
calculated.
Probability of taking
five attempts
calculated.