St Andrew’s Junior College
JC2 H2 Maths Weekend Assignment (Term 2 Week 2)
Total marks : 13
1.

Topic : Probability

Duration : 20 minutes

[RJC/2010/Mid Year] In a given population, the proportion of people that are infected with the
B3F1 virus is x. A test is conducted to determine whether a member of the population is infected.
The medical test for the infection is not completely reliable: for someone infected with the B3F1 virus, there is a probability of 0.98 that the test will prove positive; whereas for someone not infected with the B3F1 virus, there is a probability of 0.01 that the test will prove positive. One person is randomly chosen from the population and tested.
Let I be the event that the person is infected and T be the event that the person is tested positive.
Determine the least value of x such that the probability that a person is infected with the B3F1 virus given that he is tested positive is at least 0.95.
[3]
(Hint : Draw a tree diagram.)

2.

[RJC/2009/Promo]
(a) [ A' denotes the event ‘A does not occur’.]
11
8
Given that P ( A ∪ B ) = and P ( A '∩ B ) =
, determine P ( B ) if
15
45

(i)

A and B are two mutually exclusive events,

(ii)

A and B are two independent events.

[4]

(Hint for (ii) : Find P(A) first. Then use the fact that since A and B are two independent events, A' and B are independent as well.)
(Hint for both parts : Draw Venn diagram.)

(b)

In a class of 20 students, there are 7 boys and 13 girls. Of the 7 boys, 3 of them read
General Paper and of the 13 girls, 8 of them read General Paper. The rest of the students read Knowledge and Inquiry. Three students are chosen at random from the class. Find the probability that
(i)

there is at least one student who reads Knowledge and Inquiry,

(ii)

there are exactly two students who read General Paper, given that there is exactly one boy.
[4]
End of Assignment

[2]

Solutions and marking scheme for H2 Maths Weekend Assignment on Probability
1

Let I be the event that the person is infected and T be the event that the person is tested positive.

0.98

T

I

x

0.02
0.01

T

0.99

1− x

T'

T'

I'

From the probability tree diagram, we see that

P( I | T ) =
=

P( I ∩ T )
0.98 x
=
P(T )
0.98 x + 0.01(1 − x)

1 mark

98 x
.
97 x + 1

For the test to be considered reliable,

98 x
≥ 0.95
97 x + 1

1 mark (solving inequality)

Since x > 0, 97x + 1 > 0, (so we can multiply both sides by 97x + 1)

98 x ≥ 0.95(97 x + 1)
⇒ 5.85 x ≥ 0.95
⇒x≥

19
117

Least value of x =

19
(ans)
117

1 mark (end answer)

2

(i) Since A and B are two mutually exclusive events,

(a)

P ( B ) =P ( A '∩ B ) =

8
(ans)
45

1 mark (end answer)

(ii)

P ( A ) =P ( A ∪ B ) − P ( A '∩ B ) =

11 8 25 5
−
=
=
15 45 45 9

1 mark

∴ P ( A ') =1 − P ( A) =

4
9

Since A and B are two independent events, A' and B are independent as well. So we can say that P ( A '∩ B ) = P ( A ' ) × P ( B ) .

P ( A '∩ B ) = P ( A ') × P ( B ) =

8
45

1 mark

1 mark (end answer)

4
8
2
P ( B) =
⇒ P ( B ) = (ans)
9
45
5
(b)

(i) P(at least 1 student reads KI)

= 1 − P(all 3 students read GP)

 11 10 9  65
(ans)
= 1−  × ×  =
 20 19 18  76

1 mark (find

Alternative method :

1 mark (end answer)

1−

P(all read GP))

11

C3 65
=
C3 76

20

(ii) P(exactly 2 GP students given exactly 1 boy)

=

P(exactly 2 GP students and exactly 1 boy)
P(exactly 1 boy)

=

P(1 GP boy, 1 GP girl, 1 KI girl)+P(1 KI boy, 2 GP girls)
P(1 boy, 2 girls)

 3 8 5
  4 8 7 3! 
 × × × 3! +  × × × 
20 19 18
  20 19 18 2! 
=
7 13 12 3!
× × ×
20 19 18 2!
116
=
(ans)
273

Must multiply by 3! as the order of the 3 students with different criteria has to be taken into consideration.

Alternative method :

( 3C1 × 8C1 × 5C1 ) + ( 4C1 × 8C2 ) 116
=
( 7C1 × 13C2 )
273

Must multiply by

3! as the order
2!
of the 3 students where two have the same criteria has to be taken into consideration.

1 mark (find

P(1 GP boy, 1 GP girl, 1 KI girl) )
1 mark (find

P(1 KI boy, 2 GP girls) )
1 mark (find P(1 boy, 2 girls) )
1 mark (end answer)

[MJC/2006 Prelim]
A disease infects one in every 1000 people in a population. The medical test for this disease is not completely reliable: if a person has the disease, the test comes back positive 99% of the time, and if a person does not have the disease, there is a 0.5% chance of getting a positive result for the test.
(i)

Show that the probability that a person tests positive is 0.00599. Hence, find the expected number of people with positive test results in a random sample of 1700 people.
[5]

(ii)

Show that the events “has the disease” and “tested positive” are neither mutually exclusive independent. [2]

(iii)

Andrew goes for the test. Find the probability that he is infected, given that his test result is positive.
[2]

nor

Andrew’s mother decides to buy 4 different types of supplements for him. The pharmacy sells 15 different types of supplements labeled Supplement 1, 2, …, 15 respectively. Find
Let A be the event that a person has the disease.
Let B be the event that a person is tested positive.

(i)

To find P(B)
Given P(A) = 0.001, P( B | A ) = 0.99, P( B | A ' ) = 0.005, we have

P( B | A ) =

P( B | A ' ) =

P( B ∩ A)
P( B ∩ A)
⇒ 0.99 =
⇒ P( B ∩ A) = 0.00099
P( A)
0.001
P( B ∩ A ')
P( B ∩ A ')
⇒ 0.005 =
⇒ P( B ∩ A ') = 0.00495
P( A ')
0.999

P(B) = P( B ∩ A) + P( B ∩ A ') = 0.005985 = 0.00599 (to 3 s.f.)

Expected number of people tested positive = (0.005985)(1700)
= 10.1745
= 10.2 (to 3 s.f.)

(ii)

P( B ∩ A) = 0.00099 ≠ 0 . Hence, A and B are not mutually exclusive.
P(A) P(B) = (0.0001)(0.00599) = 5.99 ×10−7 ≠ P( B ∩ A) = 0.00099.
Therefore, A and B are not independent.

(iii)

To find P( A | B )
P( A | B )

(a)

(b)

P( A ∩ B ) 0.00099
=
= 0.165 (to 3 s.f.)
P( B)
0.005985

15 
 = 1365.
4

Number of ways = 

14  14  13 
 + − 
3
3
2
Required probability =       = 0.476 (to 3 s.f.)
15 
 
4

(a)

the number of ways in which she can make the purchase,

(b)

the probability that her purchase includes Supplement 1 or Supplement 2 (or both).

[1]
[2]