1

Anglo-Chinese Junior College

H2 Mathematics 9740

Qn 2012 Prelim Exam Paper 1 Solution
1(a)
( )
( )
( )
2 2 2
2
2
2 1
1 2 4 2
4 7 2 4 7 4 7
1 2
ln 4 7
2
2 3
1 2 2
ln 4 7 tan
2 3 3
x x
dx dx
x x x x x x
x x dx
x
x
x x C

+
=
+ + + + + +
= + +
+ +
+ | |
= + + +
|
\ .
} }
}


(b)
( ) ( )
( )
( )
3 3
2 2
0 0
0
5
3
2
2
0
0
5
5
2
2
2 2
3 3
2
2 2
0
3 3 5
2 2 4
0
3 5 15
a
a a
a
a
a x a x
x a xdx x dx
a x
a x dx
a
a
(

(
=
(
(

(

(
= + =
(
(

(
(
= + =
(
(

} }
}


2
( )
( )
2 2
2 2 2
2
2
2 3 2 3
0
2 5 2 2 5 2 2 5 2
1 2
0
(2 1)( 2)
1 2 0, (2 1)( 2) 0
0.5 2
, 0 0.5 2
x x x x
x x x x x x
x
x x
Since x x x
x or x
Since x x or x
+
+ + +
> >
+ + +
+ +
>

+ + > >
< >
e < < >


3









y = f(x) = 2 lnx (lnx)
2
y = 0 gives (lnx)(2 lnx) = 0
lnx = 0 or 2
x = 1 or e
2
[Note: May use GC to write e
2
correct to 3 sf since
question did not ask for exact values.]

The line 0 y = cuts y = f(x) at 2 points.
Therefore f is not a one-one function.
Thus
1
f

does not exist.

dy/dx = (2/x) 2(lnx)/x = 0 lnx = 1 x = e (so maximum point is at (e, 1))
h
1
exists when greatest possible a is e. [May write a = 2.72 (3 sf) as question did not ask for exact value.]

To find h
1
, let y = h(x) so that x = h
1
(y) where 0 < x e and y 1.
Then y = h(x) = 2 lnx (lnx)
2

Cannot factorise denominator so
cannot use Partial Fractions
method. Instead, use MF15 and
ln() since there is an x
2
in the
denominator.
LIATE method breaks down as
both terms are Algebraic.
Choose u such that the powers
of x reduce with each
differentiation and choose
something which can be
integrated for dv/dx.

Choose u = x
and dv/dx = \(a x
2
)
Inequality question with restriction on values of x,
so check domain before writing final answer.

(1,0) (e
2
,0)
0
y
x
y=f(x)
x = 0

(e, 1)

2

( )
2
ln 1 1 ln 1 1 y x x y = = but lnx 1 since 0 < x e
1 1
ln 1 1 e
y
x y x=

=
h
1
(y) =
1 1
e
y
x=

where y 1 h
1
(x) =
1 1
e
x
where x 1

Sketch graph of y = gh(x) to find
gh
R = ( 0, 3 ] e or (0, 8.15] correct to 3 sf

4 (i) Let n be the number of years after 2013.
Toms pay ( )( ) 30000 1 1500 n = + and Jerrys pay ( )
1
25000 1.05
n
=
Want ( )( ) ( )
1
30000 1 1500 25000 1.05
n
n

+ <

From the table, 17 n = .
Hence the first year in which Jerrys pay is higher than Toms is 2029.

(ii)
Toms total income ( ) ( )( ) 2 30000 1 1500
2
n
n = + (


Jerrys total income
1.05 1
25000
1.05 1
n

Want ( ) ( )( )
1.05 1
2 30000 1 1500 25000
2 1.05 1
n
n
n

+ < (




From the table, 26 n = .
Hence the first year in which Jerrys pay is higher than Toms is 2038.

5
2
2
2
2
A r B
A r B rhC h
rC
t
t t
t

= + =
( )
2
2 2 2
2 2
A r B r
V r h r A r B
rC C
t
t t t
t
| |
= = =
|
\ .

( ) ( ) ( )
2 2
2
1 1
2 0 3 0
2 2 2
gives max
3
dV r
rB A r B A r B
dr C C C
A
r V
B
t t t
t
= + = =
=

2
Cost of base
3 3
A A
r B B
B
t t
t
| |
= = =
|
\ .

May key 2013 + (X1)
into Y
3
as question
want year as final
answer. Then can read
the year from the GC
instead of calculating it
yourself.
Note that the years
form an AP.

May key 2013 + (X1)
into Y
3
as question
want year as final
answer. Then can read
the year from the GC
instead of calculating it
yourself.
Note that the years
form an AP.

3

6
1
2
3
1
4 1
3
1 1
4
2
2 1 1
1
4
3
3 3
2
2
4 1 3
1
2
9
9 9
4
6
10 1 9
1
4
27
27 27 3
10
18
28 1 27 1 3
1
10
n
n n
u
u
u
u u

= = =
+
+
= = =
+
+
= = =
+
+
= = = =
+ +
+

Let P(n) be the statement
1
1
3
1 3
n
n n
u

=
+
for all 0 n > .
When 0 n = , LHS =
0
1
4
u = RHS =
1
1
1
3 1
3
1
1 3 4
1
3

= =
+
+

LHS = RHS P(0) is true.
Assume that P(k) is true for some 0 k > , i.e.,
1
1
3
1 3
k
k k
u

=
+
.
To prove P(k + 1) is true, i.e.,
1
3
1 3
k
k k
u
+
=
+
,
( )
1
1
1
1
1
1
1 1 1 1 1
1
3
LHS
2 1
3
3
1 3
by assumption
3
2 1
1 3
3
3 3 3
1 3
RHS
2 3 1 3 2 3 1 3 1 3 3 1 3
1 3
k
k
k
k
k
k
k
k
k k k
k
k k k k k k
k
u
u
u
+

= =
+
| |
|
+
\ .
=
| |
+
|
+
\ .
+
= = = = =
+ + + + + +
+

Since P(0) is true and P(k) is true P(k+1) is true, by the Principle of Mathematical Induction,
we conclude that P(0), P(1), P(2), P(3), are all true. Hence P(n) is true for all integers 0 n > .


7
2
2 ,
t
x t y e

= =






0
x
y
(0, 1)
Either work out values yourself,
Or use GC MODE SEQ to find the
values.

Note that
the answers in the GC TABLE are
in decimals so you have to convert
them to fractions yourself,
you should re-write the expression
into u(n) =
( )
( )
3u 1 1
2u 1 1
n
n

+
first so that
it is easier to key into the GC,
you should use the BLUE u above
the 7 in your GC, not the green U.
4

2
2
x t
dx
dt
=
=

2
2
2
t
t
y e
dy
te
dt

=
=

2

t
dy
te
dx

=
Equation of normal at
( )
2
2 ,
p
p e

is ( )
2 2 1
2
p p
y e e x p
p

=
Equation of normal at
1
2, C
e
| |
|
\ .
is ( )
1
2 y e x
e
=
At ( ) , 0 A x :
2
2 2 2
1 1 2 1
2 2
e
x x
e e e

= = =
At ( ) 0, B y :
2
1 1 1 2
2 2
e
y e y e
e e e

= = =
2 2
2
1 2 1 1 2
OA
e e
e e OB e

= = : 1: e OA OB =


8
2 2 2
3 2 x a a
y x a
x a x a

= = +

so asymptotes are , x a y x a = = +
( )
2
2
d 2
1
d
y a
x
x a
= +

Since ( )
2
2
2 0 & 0 for a x a x > > e ,
( )
2
2
2
1 1 0
dy a
dx
x a
= + > =

2 2 2
3 2 x a a
y x a
x a x a

= = +

has no stationary points
(shown)








Graph of ( )
( )
2
2
2
f 1
a
y x
x a
' = = +

is
From graph of ( ) f y x ' = , f ' is increasing for x a <
OR ( )
( )
2
3
4
f 0 0
a
x x a x a
x a
'' = > < <

x a =
1 y =
x
y
x=a
y=x+a
0
(0,3a)
( )
3, 0 a

( )
3, 0 a

2 2
3 x a
y
x a

Note that the answers should
involve p only or t only, not
both p and t.
Label intersections with axes
in coordinate form (see
question).
5

9










: or
2
k k
t
t t

e < <
`
)
or
{ } ,
2
t
t t
| |

|
\ .


10
(i)
1 1 a b a b
a b a b a b a b

= =
+ +


(ii)
( ) ( )
2500
1
2500
1
2500
1
1
4 3 4 1
4 3 4 1
4 3 4 1
1
4 1 4 3
4
5
1
4
r
r
r
r r
r r
r r
r r
=
=
=
+ +
+
=
+
= +
=

1
9

+ 5
13 + 9
9997 + 9993
10001 9997
10001 1
4
| |
|
|
|
|
|
|
|
|
|
|
\ .

=

(iii)
2500 2501
0 1
2500
1
1 1
4 1 4 5 4 3 4 1
1 1
4 3 4 1 10001 10005
10001 1 1
4 10001 10005
10000 1 1
4 10001 10005
1
24.75
10001 10005
24
r r
r
r r r r
r r
= =
=
=
+ + + + +
= +
+ + +

= +
+

> +
+
= +
+
>

2
1
2
2
O
6
t


2
R
3
Re(z)
Im(z)


3
( )
3,1

Shade the region R lightly to make it
clear to the marker which is the
correct answer.

The rest of the circle and lines should
be in dashes.
Replace 10001 with a number for which
you can find the root easily and
the result is LESS than the previous
value
6

11


(i)
2
a c
OM
+
=
2
2
2
cos 60
2 2 2
0.5
2
0.5
3
( )
2 4
Shown
| | +
+ + | |
= = =
|
|
|
\ .
\ .
+
= =
+
= =
a c c
a c c a c c c
Length of Projection
c c c
a c c
Note: a c
c
c c c
c
c


11
(ii)

2
1 1 3
sin 60
4 4 8
3
8
3
8
Area of OMC
k
| |
A = = = =
|
\ .
=
=
a c a c c c Note: a c
c

11
(iii)
5
2
OD = c [Note: You may want to sketch a diagram to obtain this result.]
sin60
5 5 5 3 5 3
2 2 4 4
5 3
4
Shortest OMC OD
t

A = = = =
=
c a c a
a c a
= c
a a a a


12
2
2 4
d
2
d
d
Given , diff. w.r.t. to get
d
y
t yt
y w
t
w t
t t t

= =

2
2
2 3 3
4 4 2 2
3 2 3 2 3 2 2
2
2 2
2 2
2 2 2 2
dy
t yt
dw y y y
dt
t w t wt w t t t
dt t t t t
dy dy
t t y t y yt yt y y
dt dt
| |

|
= + = +
|
|
|
\ .
= + = +

( ) ( )
2
1
1
2
1 1 1
1
2 2 2 2
1 1
ln ln 2
2 2
1
ln ln 2
2 2 2
dy dt
y y
dy dy dt
y y y y
y y t c
y y
t c t b
y y
=
+
= =
+ +
+ = +
= + = +
+ +
} }
} } }

7

( )
2
2 2
2
2
2
2 2
e e
2
1 2 e
2 e
1 e
1 2 e
1 e
t b
t t
t
t
t
t
y
A where A
y
y Ae A
A
y
A
A
w
t A
= =
+
=
=

| |
=
|
|

\ .

When A=0, the solution is the t-axis.

13
Common ratio r = tanu . For S

to exist, 1 r < , i.e. 1 tan 1

4 4
t t
u u < < < <
For u in this range,
2 3
1
1 tan tan tan
1 tan
1 3 3
1 tan 2
2
1 tan
3 3
2
tan 1
3 3
1 3
tan
3 3
1 3 3 3
tan
3 3 3 3
2 3
tan
6
1
tan
3
6
u u u
u
u
u
u
u
u
u
u
t
u
+ + + + =

+
<

>
+
<
+
+
<
+
| |
+
<
|
|
+
\ .
<
<
<

Hence
4 6
t t
u < < .
14(a)
( )
( )
( )
( )
( )
2
2
2
2 2
2 2
2 2
2
2
2
2
3 2Re
3 2
3 4
3 6 9 0
1
2 3 0
3
1
1 4 0
3
1
1
4 12
z z
x y x
x y x
x x y
x x y
x y
x
y
+ =
+ + =
+ + =
=
=
=

=


Note: x > 0 since ( ) 3 2Re z z + =
(-1, 0) (3, 0)
x
y
3( 1) y x =
3( 1) y x =
2 2
( 1)
1
4 12
x y
=

t
w
A>0
You may want to sketch the graph of y = tanx
in your GC to find .
Use Y
1
= tanX, Y
2
= 1, Y
3
= 1
and adjust WINDOW to
X
min
= /2
X
max
= /2
X
scl
= /4
8

14(b)
( )
2 2 3 i w= +
( )
2
4 arg
3
w w
t
= =
( )
i2 / 3 i2 / 3
2 2
4 4 4 cos sin
3 3
n
n n n n
n n
w e e i
t t
t t ( | | | |
= = = +
| | (
\ . \ .

2
is real sin 0
3
3
, even, or 3 ,
2
n
n
w
n m m m n k k
t
+ +
| |
=
|
\ .
= e = e


( ) ( )
( )
( )
*
50
50 50 50
50 50
100 100 100
100
*
100 100 100 100
4 cos sin 4 cos sin
3 3 3 3
2 3
2 2 sin 2 2 2 3
3 2
2 3
w w w w
i i
i i i
k
t t t t
t
=
( ( | | | | | | | |
= +
| | | | ( (
\ . \ . \ . \ .
( | |
( | |
= = =
( |
| (
|
\ . (
\ .
=




Qn 2012 Prelim Exam Paper 2 Solution
1 (i)













(ii)
2.5 3 4
3 0.5 7
2 0.5 5
, , ,
2.5 4 3
0.5 3 7 0
1.5 5 0
, :
k
k k
k k
Equating components
k
k
k
Solving Equations k






+ + | | | |
| |
+ = +
| |
| |
+ +
\ . \ .
+ =
+ =
+ =
=
i j k
---(1)
---(2)
---(3)
(1) (2), (3) 2 ( ) . 1 2. = = ,


( )
( )( )
( )( )
( )
2.5 (2) 2 7
2, 2, 2 3 2 8
2 2 2 6
7,8, 6
When k
+ | | | |
| |
= = = + =
| |
| |
+
\ . \ .
r
Intersection point is


2









Let ( ) y f x =
1
sin 2 1
2
2
ln sin 2
1 2
diff. w.r.t. : 1 4 2
1 4
x
y e y x
dy dy
x x y
y dx dx
x

= =
= =

2
2
2
2
4
diff. w.r.t. : 1 4 2
1 4
x dy d y dy
x x
dx dx dx
x

+ =

Use exponential form for indices.
Use Cartesian or trigonometric
for real and imaginary parts.
9

( )
( )
( )
( )
2
2 2
2
2
2
2
2
2
2
4 1 4 2 1 4
1 4 4 2 2
1 4 4 4 (shown)
dy d y dy
x x x
dx dx dx
d y dy
x x y
dx dx
d y dy
x x y
dx dx
+ =
= +
= +

( )
( ) ( )
3 2 2
2
3 2 2
diff. w.r.t. : 1 4 8 4 4 4
d y d y d y dy dy
x x x x
dx dx dx dx dx
+ = + +

When x = 0, ( ) ( ) ( ) ( ) 0 1, ' 0 2, '' 0 4, ''' 0 16 f f f f = = = =
( )
2 3
8
1 2 2 ...
3
f x x x x = + + + +

1
sin 2 1
6
2 3
6
sin 2
6
1 1
2
2 4
1 1 8 1 2
1 2 2 1
4 4 3 4 3
x
e e x
x x
e

t
t
= =
= =
| | | | | |
~ + + + =
| | |
\ . \ . \ .


3i













ii
a)













1 sin
diff. w.r.t. : cos
x u
dx
x u
du
= +
=


( ) ( )
( )
( )
2
2
2
2 2
2 1
2 2 1 sin 1 sin cos
2 2sin 1 2sin sin cos
cos 2 1
1 sin cos cos
2
1
sin 2 1 2sin cos 1 1
2 sin 1
4 2 4 2 2 2
x x dx u u u du
u u u u du
u
u u du u du du
x
u u u
u c u c x x x c

= + +
= +
+
= = =

= + + = + + = + +
} }
}
} } }




( )
1 1 1 1
9
2 2 9
0 0 0 0
1
2 10 2
2 1
0
Area =2 2 2 =2 2 2
1 1 1 19
2 2 2 sin 1 2 2
2 2 2 10 2 10 4 5 2
R x x x dx x x dx x x x dx x xdx
x x x x
x x x x

( (
+ + + +
( (

( t t (
= + + = + =
(
(


} } } }



10

b)


( )
( )
2
1 1 2
2 9
0 0
Vol. of revolution formed when R is rotated completely about -axis
2 2 2 14.995
x
x x x dx x x dx
(
= t + + =
(

} }


4 (i)




















4(ii)
Method 1:
1 1
: 1 2
0 3
r line l
| | | |
| |
= +
| |
| |

\ . \ .

Since (1,1,0) lies on
2
p ,
1 1
1 1 1 1 2
0 1
a a a
| | | |
| |
= + = =
| |
| |
\ . \ .

Method 2:
Since direction vector of line l is to normal of
2
p ,
1 1
2 0 1 2 3 0 2
3 1
a a a
| | | |
| |
= + = =
| |
| |

\ . \ .

Method 3:
1 1 1
1 2 1 1 2
1 1 1
a
a a a
a
| | | | | |
| | |
= = =
| | |
| | |

\ . \ . \ .


Method 1: Normal method
1
1 0 1
: 1 2 : 0 1
1 1 1
0 1
1
0 1 2 1 1
3
1 1
p and line AN

| | | | | |
| | |
= = +
| | |
| | |
\ . \ . \ .
+ | | | |
| |
+ = + + + = =
| |
| |
+
\ . \ .
r r

0 1 1
1 1
0 1 1
3 3
1 1 4
ON
| | | | | |
| | | | | | |
= + =
| |
| | |
\ . \ .
| | |
\ . \ . \ .


Method 2: Projection Method
1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
. 1 1 1 1 1 1 1
3 3 3 3 3 3
1 1 1 1 1 1 1
AN AB
| || | | || | | | | | | | | | | | | | | |
+ | | | | | | | | | | | | |
= = = =
|
| | | | | | | | | | |
\ .
| | | | | | | | | | |

\ . \ . \ . \ . \ . \ . \ . \ .\ . \ .\ .

1 0 1
1 1
1 0 1
3 3
1 1 4
AN ON OA
ON AN OA
=
= +
| | | | | |
| | |
= + =
| | |
| | |
\ . \ . \ .




11

4
(iii)
Let the acute angle between planes
1
p and
2
p be u .
1 1
1 2
1 1
1 2 1 2
cos
1 1 3 6 18
1 2
1 1
61.9
u
u
| | | |
| |
| |
| |
+ +
\ . \ .
= = =
| | | |
| |
| |
| |
\ . \ .
=


4
(iv)
Distance from Origin to p
1
2
2
1 3
1
1
= =
| |
|
|
|
\ .


Distance from Origin to p
3
( 0)
2 12
2
2

b
b
b = = >
| |
|
|
|
\ .

2
3
12 3
2
3
12 3
4 6 10
b
b
b
=
= +
= + =


4
(v)


















(vi)
1
p ,
2
p and
4
p meet in a line l.
Hence (1,1,0) lies on
4
p .
1
1 2
0 3
2
c
d
c d
| | | |
| |
=
| |
| |
\ . \ .
+ =


Normal of
4
p is perpendicular to line l.
1
2 2 0
3 3
4 9 0
5
7
c
c
c
d
| | | |
| |
=
| |
| |

\ . \ .
+ =
=
=


5 , 7 c d = =




12

5 Randomly choose 10 programmers, 1 secretary and 1 section head for the sample.

Stratified sampling guarantees a representative sample of each group (i.e. programmers,
secretaries and section heads) in the population.
CANNOT accept any of the following answers:
+ allows the opinions of different strata to be considered separately.
+ accurate
+ unbiased

6
(a)
(i)




(ii)

(b)
4 boys and 3 girls

_ B _ B _ B _ B _ 4!
5
P
3
= 24 60
= 1440

Or 4! (
5
C
3
3!) = 24 60 = 1440

G BBBB GG type so 4! 4! = 576

Case (i): 3 boys and 3 girls

4
C
3
(3 1)! 3!
= 4 2 6 = 48


Case (ii): 4 boys and 2 girls
(4 1)!
3
C
2

4
P
2

= 6 3 12
= 216

Or (4 1)!
3
C
2

4
C
2
2! = 216

Total number of arrangements = 48 + 216 = 264

7 Plot of y against x.
Highest y = 6.3; Lowest y = 2.7
Highest x = 12.3; Lowest x = 4.4
Correlation coefficient = r = 0.263

A linear model is not appropriate as the scatter diagram
shows that the points are not close to a straight line and
the value of r is quite close to 0.


7

For y = ax
2
+ b, value of r = 0.913
For y = a ln x + b, value of r = 0.971
Therefore, (b) y = a ln x + b is a better model as the value of r in (b) is closer to 1 than for (a).

Line of regression is y = 11.042 948 39 2.687 231 256 ln x
When y = 6.1, ln x = 1.839 420 548 x = 6.29 (3 sf) Accept x = 6.3 (1 dec place)

This estimate is valid since the value of r is close to 1 and the value of y used in within the
range of experimental data (4.5 y 6.3).
Or, may say that x comes from interpolation instead of within the range of experimental data.

B
B
B G
G
G
B
B
B G
B
G
x
y
4.4
12.
3
2.7
6.3
13

8(a) Let X be the number of cars arriving at the jetty in 30 mins. X ~ Po(6)

( ) ( ) ( ) 2 0.0446 ( ) 2 0.0620 i P X ii P X = = s =
8(b) Let Y be the number of cars arriving at the jetty in 20 mins. Y ~ Po(4)

( )
( )
0.1
0.9
7
P Y k
P Y k
Least k
> <
s >
=

k
( ) P Y k s
6 0.88933
7 0.94887
8 0.97864

9(a) Let X be the time of journey from Town A to Town B.
( )
( ) ( )
2
Let ,
1 1
60 70
4 20
60 1 70 1
4 20
60 70
0.67449 1. 64485
Solve: 53.0491 53
10.3054 10 (nearest minute)
X N
P X P X
P Z P Z
o

o o

o o

o
> = > =
| | | |
> = > =
| |
\ . \ .

= =
= ~
= ~


9(b) Let Y be the time of journey from Town B to Town C.

( )
( )
80, 9
0.02
86.1612
Min 87mins
Last time of departure 10 33h
Y N
P Y k
k
k
> s
>
=


( ) ( )
9
80,
20
80 2 78 82 0.997
Y N
P Y P Y
| |
|
\ .
< = < < =


10

Let L be the length and B be the breadth of a tile
( ) ( )
( )
( )
( )
( )
2 2
2 2
1 2 3 10
1 2 3 10
18.9,0.3 8.9, 0.1
2 2
2(18.9) 2(8.9) 55.6
4(0.3 ) 4( 0.1 ) 0.4
............ 10(55.6) 556
............ 10(0.4) 4
L N B N
Let P L B
E P
Var P
E P P P P
Var P P P P
= +
= + =
= + =
+ + + + = =
+ + + + = =






14

10 Let W be the number of tiles with a red tint. W ~ B(500, 0.6)
( )
( )
( )
300 5 200 5, is large 300,120
0.95
0.05
1
0.05
2
1
281.98
2
282.48
Greatest =282
np and nq n W N approximately
P W k
P W k
P W k
k
k
k
= > = >
> >
< s
| |
< s
|
\ .
<
<


11
( )
0
1
0
0
To test : H : 27
H : 27 at 2%level
27
Under H : 26
27
Test statistic = 1.22474
p-value = 0.1158 >0.02
Do not reject H
X
T t
s

=
>

=


There is insufficient evidence at the 2% level of significance to conclude the mean foot length
of an 18 year old man of high intelligence is more than 27cm

Assumption: Assume the population of foot length of 18 year old man of high intelligence is
normally distributed


0
1
0
2
2
To test : H :
H : at 4%level
Under H by Central limit theorem, since is large
1
, , (123.20) 2.088136
60 59
k
k
n
s
X N k approx where s

=
=
| |
= =
|
\ .

0
Toreject H : 2 0.04 0.02
60 60
2.05375 2.05375
60 60
26.6 26.6
2.05375 2.05375
2.088136 2.088136
60 60
26.9831 26.21687
27.0 26.2
x k x k
P Z P Z
s s
x k x k
or
s s
k k
or
k or k
k or k
| | | |

| |
< < < <
| |
| |
\ . \ .

< >

< >
> <
> <





15

12
(i)
Let X be the r.v. number of shots, out of 20, that hit the target. X ~ B(20, 0.25)

Most likely number of shots is 5.

12
(ii)
P(X 5 | X 10) =
( )
( )
P 5 10
P 10
X
X
s s
s
=
( ) ( )
( )
P 10 P 4
P 10
X X
X
s s
s
=
0.5812163575
0.9960578583
= 0.584 (3 sf)

12
(iii)
P(same number of shots each to hit target)
= P(1, 1) + P(2, 2) + P(3, 3) +
= 0.25
2
+ (0.75 0.25)
2
+ (0.75 0.75 0.25)
2
+
= 0.25
2
(1 + 0.75
2
+ 0.75
4
+ 0.75
6
+ )
= 0.0625
2
1
1 0.75

= 0.0625
1
0.4375
=
1
7
or 0.143 (3 sf)