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CHAPTER 7 PROBABILITY FORM 5
PAPER 1
1. A box contains 8 blue marbles and f white marbles. If a marble is picked randomly from the
box, the probability of getting a white marble is .
Find the value of f.[3 marks]
2. A bag contains 4 blue pens and 6 black pens. Two pens are drawn at random from the bag one after another without replacement. Find the probability that the two pens drawn are of different colours.
[3 marks]3. Table 2 shows the number of coloured cards in a box.
Colour Number of CardsYellow 6Green 4Blue 2
Table 2
Two cards are drawn at random from the box.Find the probability that both cards are of the same colour.
[3 marks]
4. The probability that Amir qualifies for the final of a track event is while the probability
that Rajes qualifies is .
Find the probability that,(a) both of them qualify for the final,(b) only one of them qualifies for the final. [3 marks]
5. A box contains 4 cards with the digits 1, 2, 3 and 4. Two numbers are picked randomly from the box. Find the probability that both the numbers are prime numbers.
[3 marks]
6. Team A will play against Team B and Team C in a sepak takraw competition. The
probabilities that Team A will beat Team B and Team C are and respectively. Find the
probability that Team A will beat at least one of the teams.
[3 marks]
120
CHAPTER 7 PROBABILITY FORM 5
7. The probability of a particular netball player scoring a goal in a netball match is . Find the
probability that this player scores only one goal in three matches.[3 marks]
8. The probabilities that Hamid and Lisa are selected for a Science Quiz are and
respectively. Find the probability that at least one of them are selected.[3 marks]
9. Table 9 shows the number of coloured marbles in a box.
Colour Number of MarblesBlack 2Red 5
Green 3Yellow 8
Table 9
Two marbles are drawn at random from the box.Find the probability that both marbles are of the same colour.
[3 marks]
10. The probability of Zainal scoring a goal in a football match is . Find the probability that
Zainal scores at least one goal in two matches.[3 marks]
121
CHAPTER 7 PROBABILITY FORM 5
ANSWER (PAPER 1)
No. Solution Marks
1. 8 + f 1
f
8 + f =
3
5
1
f = 12 1
2. = 4
10 x
6
9 @
6
10 x
4
9 1
= 4
10 x
6
9 +
6
10 x
4
9 1
= 8
151
3. = 6
12 x
5
11 or
4
12 x
3
11 or
2
12 x
1
11 1
= 6
12 x
5
11 +
4
12 x
3
11 +
2
12 x
1
11 1
= 1
31
4(a) =
2
15
1
(b) = 1
5 x
1
3 +
2
3 x
4
5 1
122
CHAPTER 7 PROBABILITY FORM 5
= 3
51
5. =
2
4 or
1
3
1
= 2
4 x
1
3 1
= 1
61
6. =
1
3 x
3
5 o r
2
3 x
2
5 or
1
3 x
2
5 1
= 1
3 x
3
5 +
2
3 x
2
5 +
1
3 x
2
5 1
= 3
51
7. = 1
5 x
4
5 x
4
5 1
= 1
5 x
4
5 x
4
5 +
4
5 x
1
5 x
4
5 +
4
5 x
4
5 x
1
5 1
= 48
125 1
8. = 2
5 x
2
3 or
1
3 x
3
5 or
1
3 x
2
5 1
123
CHAPTER 7 PROBABILITY FORM 5
= 2
5 x
2
3 +
1
3 x
3
5 +
1
3 x
2
5 1
= 3
5 1
9. = 2
18 x
1
17 or
5
18 x
4
17 or
3
18 x
2
17 or
8
18 x
7
17 1
= 2
18 x
1
17 +
5
18 x
4
17 +
3
18 x
2
17 +
8
18 x
7
17 1
= 14
51 1
10. =
2
5 x
2
5 or
2
5 x
3
5 or
3
5 x
2
5 1
= 2
5 x
2
5 +
2
5 x
3
5 +
3
5 x
2
5 1
= 16
25 1
124
