Evaluation of Mathematics

1. A bag contains ten counters, of which six are red and four are green. A counter is chosen at

random, its colour is noted and it is replaced in the bag. A second counter is then chosen at

random. Find the probabilities that :

a. both counters are red

b. both counters are green

c. just one counter is red

d. at least one counter is red

e. the second counter is red

2. Events A, B, and C satisfy these conditions:

P(A) = 0.6; P(B) = 0.8; P(B|A) = 0.45; P(B and C) = 0.28

Calculate :

a. P(A and B)

b. P(C|B)

c. P(A|B)

3. Calculate the number of arrangements of the letter in the word NUMBER. How many

arrangements if at begin and end is a vowel ?

4. a. Given that 5 sin X + 12 cos X = R sin ( X + Y ),

find the values of X and Y for which R > 0 and 0 < 90.

b. Find all values of X between 0 and 360 satisfying :

(i) 5 sin X + 12 cos X = 4

(ii) 5 sin 2X + 12 cos 2X = 13 sin X

c. Find the greatest and smallest positive value of X at which this occurs:

12 cos X + 5 sin X + 20

Solution:

1. Red Counter = 6Green Counter = 4

a.C16

C110 ×

C16

C110=

610×610

= 36100

= 925

b. C14

C110 ×

C14

C110=

410×410

= 16100

= 425

c. C16

C110 ×

C14

C110+

C14

C110 ×

C16

C110=

610×410

+ 410×610

=2410

+ 2410

= 48100

=1225

d. C16

C110 ×

C14

C110+

C14

C110 ×

C16

C110+

C16

C110 ×

C16

C110

¿ 610×410

+ 410×610

+ 610×610

¿ 2410

+2410

+ 3610

= 84100

=2125

e. C14

C110 ×

C16

C110=

410×610

= 24100

= 625

2. P(A) = 0.6

P(B) = 0.8

P(B|A) = 0.45

P(B and C) = 0.28

a. P(A ∩ B) = P(A) x P(B|A)

= 0.6 x 0.45

= 0.27

b. P(B ∩ C ) = P(B) x P(C|B)

0.28 = 0.8 x P(C|B)

0.35 = P(C|B)

c. P(B ∩ A) = P(B) x P(A|B)

P(A ∩ B) = 0.8 x P(A|B)

0.27 = 0.8 x P(A|B)

0.3375= P(A|B)

3. Vowel Letter { U and E } = 2

Consonant Letter { N, M, B, R } = 4

With Filling Slot we can find that arrangements if at begin and end is a vowel

2 4 3 2 1 1

So there are 2 x 4 x 3 x 2 x 1 x 1 = 48 ways4. a. r > 0

5 sin X + 12 cos = r sin (X + Y)= r (sin X cos Y + sin Y cos X)= r cos Y sin X + r sin Y cos X

- r cos Y = 5

- r sin Y = 12

tan Y = sinYcosY

= 125

Y = arc tan 125

= 67,38

Then we can find the value of R,

R2 . sin2Y + R2 . cos2Y = 144 + 25

R2 (sin2Y + cos2Y ) = 169

R2 = 169

R = + 13

R > 0 so R = 13

For all Real of X, 5 sin X + 12 cos X always equal with 13 sin ( X + Y )

b. (i) 5 sin X + 12 cos X = 4k . sin ( X + Y ) = 413 sin ( X + Y ) = 4

sin ( X + Y ) = 413

sin ( X + Y ) = sin 17,92= sin ( 180 - 17,92 )

So,X + Y = 17,92……………………………..X + Y = 180 - 17,92…………………….

(ii) 5 sin 2X + 12 cos 2X = 13 sin X

13 sin ( X + Y ) = 13 sin X

sin ( X + Y ) = sin X

= sin ( 180 - X )

So,

X + Y = X………………….……………..

X + Y = 180 - X…………….……….….

c. 12 cos X + 5 sin X + 20{ k . sin ( a + X ) } = k ( sin a . cos X + cos a . sin X )

= k . sin a . cos X + k . cos a . sin X= 12 cos X + 5 sin X

So that,k . sin a = 12k . cos a = 5Then we can find the value of k,k2 . sin2a + k2 . cos2a = 144 + 25k2 (sin2a + cos2a ) = 169k2 = 169 => k = + 13

12 cos X + 5 sin X + 20= { k . sin ( a + X ) } + 20

the greatest positive value, so sin ( a + X ) = 1 and k = 13= { k . sin ( a + X ) } + 20= 13 + 20= 33the smallest positive value, so sin ( a + X ) = 1 and k = -13 , or sin ( a +

X ) = -1 andk = 13= { k . sin ( a + X ) } + 20= -13 + 20= 7

Finished by

Name : Fandy Ahmad

Grade : XI.IA 8

Absent : 11