7.5 SPM Practice (Long Questions)


Question 10 (5 marks):
The table below shows some members of Red Crescent Society and St John Ambulance Society that are instructed to do an outdoor tasks at various places during the Flag Day.


Two members from the societies were dropped off random at those places.
(a) List all the possible outcomes of the event in this sample space.
You also may use the letters such as A for Amy and so on.

(b) By listing down all the possible outcomes of the event, find the probability that
(i) a boy and a girl were dropped off at a certain place.
(ii) both members that were dropped off at a certain place are from the same society.


Solution:
(a)
S = {(A, J), (A, N), (A, F), (A, C), (A, M), (J, A), (J, N), (J, F), (J, C), (J, M), (N, A), (N, J), (N, F), (N, C), (N, M), (F, A), (F, J), (F, N), (F, C), (F, M), (C, A), (C, J), (C, N), (C, F), (C, M),(M, A), (M, J), (M, N), (M, F), (M, C)}

(b)(i)
{(A, F), (A, C), (A, M), (J, F), (J, C), (J, M), (N, F), (N, C), (N, M), (F, A), (F, J), (F, N), (C, A), (C, J), (C, N), (M, A), (M, J), (M, N)}
Probability = 18 30 = 3 5

(b)(ii)
{(A, F), (A, C), (F, A), (F, C), (C, A), (C, F), (J, N), (J, M), (N, J), (N, M), (M, J), (M, N)}
Probability = 12 30 = 2 5

7.5 SPM Practice (Long Questions)


Question 9 (6 marks):
Diagram 10.1 shows a disc with four equal sectors and a fixed pointer. Each sector is labelled with water heater, oven, television and iron respectively. Diagram 10.2 shows a box which contains three cash vouchers, RM10, RM20 and RM50.



A lucky customer in a supermarket is given a chance to spin the disc once and then draw a cash voucher from the box.
(a) List the sample space for the combination of prizes that can be won.

(b)
By listing down all the possible outcomes of the event, find the probability that

(i)
 the customer wins a television or cash voucher worth RM50,

(ii)
 the customer does not wins the water heater and the cash voucher worth RM20.


Solution:
(a)
S = {(Water heater, RM10), (Water heater, RM20), (Water heater, RM50), (Oven, RM10), (Oven, RM20), (Oven, RM50), (Television, RM10), (Television, RM20), (Television, RM50), (Iron, RM10), (Iron, RM20), (Iron, RM50)}

(b)(i)
{( Water heater, RM50), (Oven, RM50), (Iron, RM50), (Television, RM10), (Television, RM20), (Television, RM50)}
Probability = 6 12 = 1 2

(b)(ii)
Probability =1P( Water heater, RM20 ) =1 1 12 = 11 12

7.5 SPM Practice (Long Questions)


Question 7:
Diagram below shows 4 cards labelled with letters.


Two cards are chosen at random from the box, one by one, without replacement.
(a) List the sample space.
(b) Find the probability that
(i) at least one card chosen is labelled C,
(ii) both cards chosen are labelled with same letter.

Solution:
(a)
Sample space, S
= {(C, O1), (C, O2), (C, L), (O1, C), (O1, O2), (O1, L), (O2, C), (O2, O1), (O2, L), (L, C), (L, O1), (L, O2)}
n(S) = 12

(b)(i)
At least one card chosen is labelled C
= {(C, O1), (C, O2), (C, L), (O1, C), (O2, C), (L, C)}
Probability= 6 12 = 1 2

(b)(ii)
Both cards chosen are labelled with same letter
= {(O1, O2), (O2, O1)}
Probability= 2 12 = 1 6


Question 8 (6 marks):
A bag contains five cards, labelled with the letters I, J, K, M and U.
One card is picked at random from the bag and the letter is recorded. Without replacement, another card is picked at random from the bag and the letter is also recorded.

(a)
 Complete the sample space in the answer space.

(b)
By listing all the possible outcomes of the event, find the probability that

(i)
 the first card picked is labelled with a vowel.

(ii)
 the first card picked is labelled with a consonant and the second card picked is labelled with a vowel.

Answer:
{(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), (   , ), (   , ), ( , ), ( , ), (   , ), (   ,   ), (   , ), (   ,   ), (   , ), (   ,   )}


Solution:
(a)
S = {(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), (K, M), (K, U), (M, I), (M, J), (M, K), (M, U), (U, I), (U, J), (U, K), (U, M)}

(b)(i)
{(I, J), (I, K), (I, M), (I, U), (U, I), (U, J), (U, K), (U, M)}
Probability of first card with vowels = 8 20 = 2 5

(b)(ii)
{(J, I), (J, U), (K, I), (K, U), (M, I), (M, U)}
Probability = 6 20 = 3 10

7.5 SPM Practice (Long Questions)


Question 5:
A fair dice is tossed. Then a card is picked at random from a box containing a yellow, a red and a purple card.

(a) By using the letter Y to represent the yellow card, the letter R to represent the red card and the letter P to represent the purple card, complete the sample space in the diagram in the answer space.

(b) By listing down all the possible outcomes of the event, find the probability that
(i) a number less than six and a red card are chosen.
(ii) a number greater than three or a purple card is chosen.

Answer:

(a)

Solution:
(a)


(b)
( i ) Sample space of a number less than six and a red card are chosen { ( 1,R ),( 2,R ),( 3,R ),( 4,R ),( 5,R ) } Probability= 5 18 ( ii ) Sample space of a number greater than three or a purple card is chosen {( 4,Y ),( 5,Y ),( 6,Y ),( 4,R ),( 5,R ),( 6,R ),( 4,P ),( 5,P ),( 6,P ) ,( 1,P ),( 2,P ),( 3,P )} Probability= 12 18  = 2 3



Question 6:
A coin is tossed and a dice is rolled simultaneously. By listing the sample of all the possible outcomes of the event, find the probability that
(a) a head and an odd number are obtained.
(b) a head or a number greater than 4 are obtained.

Solution:
( a ) Sample space of a head and an odd number { ( H,1 ),( H,3 ),( H,5 ) } Probability= 3 12   = 1 4 ( b ) Sample space of a head or a number greater than 4 { ( H,1 ),( H,2 ),( H,3 ),( H,4 ),( H,5 ),( H,6 ),( T,5 ),( T,6 ) } Probability= 8 12   = 2 3

7.5 SPM Practice (Long Questions)


7.5.2 Probability (II), SPM Practice (Long Questions)
Question 3:
Diagram below shows two cards labelled with letters in box A and three numbered cards in box B.


A card is picked at random from box A and then a card is picked at random from box B.
By listing the sample of all possible outcomes of the event, find the probability that

(a) a card labelled M and a card with an even number are picked,
(b)   a card labelled Q or a card with a number which is multiple of 2 are picked.

Solution:
Sample space, S
= {(M, 2), (M, 3), (M, 6), (Q, 2), (Q, 3), (Q, 6)}
n(S) = 6

(a)
{(M, 2), (M, 6)}
P( M and even number )= 2 6 = 1 3  
 
(b)
{(Q, 2), (Q, 3), (Q, 6), (M, 2), (M, 6)}
P( Q or multiple of 2 )= 5 6



Question 4:
Table below shows the names of participants from two secondary schools attending a public speaking training programme.
Boys
Girls
School A
Karim
Rosita
Sally
Linda
School B
Ahmad
Billy
Nancy
Two participants are required to give speeches at the end of the programme.
(a)  A participant is chosen at random from School A and then another participant is chosen at random also from School A.
(i) List all the possible outcomes of the event in this sample space.
(ii)Hence, find the probability that a boy and a girl are chosen.

(b)  A participant is chosen at random from the boys group and then another participant is chosen at random from the girls group.
(i) List all the possible outcomes of the event in this sample space.
(ii)Hence, find the probability that both participants chosen are from School B.

Solution:
(a)(i)
Sample space, S
= {(Karim, Rosita), (Karim, Sally), (Karim, Linda), (Rosita, Sally), (Rosita, Linda), (Sally, Linda)}
n(S) = 6

(a)(ii)
{(Karim, Rosita), (Karim, Sally), (Karim, Linda}
P ( a boy and a girl ) = 3 6 = 1 2
 
(b)(i)
Sample space, S
= {(Karim, Rosita), (Karim, Sally), (Karim, Linda), (Karim, Nancy), (Ahmad, Rosita), (Ahmad, Sally), (Ahmad, Linda), (Ahmad, Nancy), (Billy, Rosita), (Billy, Sally), (Billy, Linda), (Billy, Nancy)}
n(S) = 12

(b)(ii)
{(Ahmad, Nancy), (Billy, Nancy)}
P (both participants from School B)
= 2 12 = 1 6

7.4 SPM Practice (Short Questions)


Question 6:
A box contains 18 black cards and 8 blue cards. Fanny puts another 2 black cards and 4 blue cards inside the box. A card is chosen at random from the box.
What is the probability that a black card is chosen?

Solution:
New total number of cards in the box =( 18+2 )+( 8+4 ) =32 Probability a black card is chosen Probability= 18+2 32   = 20 32   = 5 8


Question 7:
A fruits seller wants to check the number of rotten apples in each box. Table below shows the number of rotten apples in each box.
Given that there were 55 boxes of apples.
If a box is selected at random, what is the probability that the box does not contain rotten apples?

Solution:
Number of boxes without rotten apples =55( 3+9+2+3+1+2 ) =5520 =35 Probability= 35 55 = 7 11


Question 8:
A form 5 science class has 15 boy students and a number of girl students. A student is chosen at random from the class. The probability of choosing a boy student is 3 5 .
Find the number of girl students in the class.

Solution:
Let x be the number of girl students in the class. P( a girl student )=1 3 5   = 2 5 x 15+x = 2 5 5x=30+2x 3x=30 x=10

7.4 SPM Practice (Short Questions)


Question 4:
A box contains 5 red cards, 3 yellow cards and a number of green cards. A card is picked at random from the box. Given that the probability of picking a yellow card is 1 6 , find the probability of picking a card that is not green.

Solution:
P( yellow card )= n( yellow card ) n( S )                       1 6 = 3 n( S )                 n( S )=3×6                         =18 n( not green card )=5+3=8 P( not green card )= 8 18                             = 4 9



Question 5:
In an office, 300 workers go to work by bus, 60 workers go by car and the rest go by motorcycle. A worker is chosen at random. The probability of choosing a worker who goes to work by bus is .
Find the probability of choosing a worker who does not go to work by motorcycle.

Solution:
Let y be the number of workers who go to office by motorcycle. P( by bus )= 2 3 Number of workers ( By bus ) Total number of workers = 2 3 300 300+60+y = 2 3 900=720+2y 2y=180 y=90 Probability= 300+60 360+90  = 360 450  = 4 5

7.2 Probability of the Complement of an Event


7.2 Probability of the Complement of an Event

1. The complement of an event A is the set of all the outcomes in the sample space that are not included in the outcomes of event A.

2. The probability of the complement of event A is:
P ( A ' ) = 1 P ( A )

Example:
A number is chosen at random from a set of whole number from 1 to 40. Calculate the probability that the chosen number is not a perfect square.

Solution:
Let
A = Event of choosing a perfect square.
A’ = Event that the number is not a perfect square.
A = {1, 4, 9, 16, 25, 36}
n(A) = 6
P ( A ) = n ( A ) n ( S ) = 6 40 = 3 20 P ( A ' ) = 1 P ( A ) = 1 3 20 = 17 20 Hence, the probability that the number chosen is not a perfect square is 17 20 .

7.4 SPM Practice (Short Questions)


Question 1:
A bag contains 36 marbles which are black and white. It is given that the probability for a black marble being picked at random from the bag is 5 9 .
Calculate the number of white marbles to be taken out from the bag so that the probability of picking a black marble is   5 8 .

Solution:

Number of black marbles in the bag = 5 9 ×36=20 Let y is the total number of marbles left in the bag. y× 5 8 =20 y=20× 8 5 =32 Number of white marbles to be taken out from the bag  = 3632 =4



Question 2:
Table below shows the number of different coloured balls in three bags.

Green
Brown
Purple
Bag A
3
1
6
Bag B
5
3
4
Bag C
4
6
If a bag is picked at random and then a ball is drawn randomly from that bag, what is the probability that a purple ball is drawn?

Solution:

Probability of picking a bag = 1 3
Probability of picking purple ball from bag A = 6 10 = 3 5
Probability of picking purple ball from bag B = 4 12 = 1 3
Probability of picking purple ball from bag C = 2 12 = 1 6

P ( purple ball ) = ( 1 3 × 3 5 ) + ( 1 3 × 1 3 ) + ( 1 3 × 1 6 ) = 1 5 + 1 9 + 1 18 = 11 30


Question 3:
A box contains 48 marbles. There are red marbles and green marbles. A marble is chosen at random from the box. The probability that a red marble is chosen is 1 6 .
How many red marbles need to be added to the box so that the probability that a red marble is chosen is  1 2 .

Solution:
Number of red marbles in the box = 1 6 ×48 =8 Let the number of red marbles needed to be added be y. P( red marble )= 1 2 8+y 48+y = 1 2 16+2y=48+y 2yy=4816 y=32 Number of red marbles need to be added =32

7.3 Probability of a Combined Event


7.3 Probability of a Combined Event
 
7.3b Finding the Probability of Combined Events (a) A or (b) A and B
1. The probability of a combined event ‘A or B’ is given by the formula below.
  P(A or B)=P(AB)     = n(AB) n(S)     
2. The probability of a combined event ‘A and B’ is given by the formula below.
  P(A and B)=P(AB)      = n(AB) n(S)   

Example:
The probabilities that two Form 5 students, Fiona and Wendy will pass the English oral test are 1 3  and  2 5  respectively. Calculate the probability that
(a) both Fiona and Wendy past the English oral test,
(b) both Fiona and Wendy fail the English oral test,
(c) either one of them passes the English oral test,
(d) at least one of them passes the English oral test.  

Solution:
Let
F = Event that Fiona passes the English oral test
W = Event that Wendy passes the English oral test
Therefore,
F’ = Event that Fiona fails the English oral test
W’ = Event that Wendy fails the English oral test
P ( F ) = 1 3 , P ( F ' ) = 2 3 P ( W ) = 2 5 , P ( W ' ) = 3 5

(a)
P (both Fiona and Wendy past the English oral test)
= P (F W)
= P (F) x P (W)
= 1 3 × 2 5 = 2 15

(b)
P (both Fiona and Wendy fail the English oral test)
= P (F’ W’)
= P (F’) x P (W’)
= 2 3 × 3 5 = 2 5

(c)
P (either one of them passes the English oral test)
= P (F W’) + P (F’ W)
= (P (F) x P (W’)) + (P (F’) x P (W))
= ( 1 3 × 3 5 ) + ( 2 3 × 2 5 ) = 7 15

(d)
P (at least one of them passes the English oral test)
= 1 – P (Both of them fail) ← (concept of complement event)
= 1 – P (F’) x P (W’)
= 1 2 5 = 3 5