Question 1
10 counters are in a bag, 6 are blue and 4 are yellow.
One counter is taken from the bag at random and not replaced.
A second counter is then taken from the bag at random.
1(a) Complete the probability tree diagram below showing the probabilities of taking counters from the bag.
ANSWER: Multiple Choice (Type 1)
A: A=\dfrac{6}{10}, B=\dfrac{5}{10}, C=\dfrac{4}{10}, D=\dfrac{6}{9}
B: A=\dfrac{6}{10}, B=\dfrac{5}{9}, C=\dfrac{4}{9}, D=\dfrac{6}{9}
C: A=\dfrac{6}{10}, B=\dfrac{6}{10}, C=\dfrac{4}{10}, D=\dfrac{6}{9}
D: A=\dfrac{6}{10}, B=\dfrac{6}{10}, C=\dfrac{4}{10}, D=\dfrac{6}{9}
Answer: B
Workings:
Probabilities add up to 1 for each event. The first counter is not replaced, so there are 9 counters in total in the second event (i.e. all probabilities are expressed as 9^{\text{th}}s. So,
A=1-\dfrac{4}{10}=\dfrac{6}{10}
B=\dfrac{5}{9}
C=\dfrac{4}{9}
D=\dfrac{6}{9}
Marks = 2
1(b) Calculate the probability that after the second pick, 1 blue counter and 1 yellow counter has been removed from the bag.
Give your answer as a fraction in its simplest form.
ANSWER: Fraction
Answer: \dfrac{8}{15}
Workings:
\text{P(Blue then Yellow)}=\dfrac{6}{10}\times\dfrac{4}{9}=\dfrac{4}{15}
\text{P(Yellow then Blue)}=\dfrac{4}{10}\times\dfrac{6}{9}=\dfrac{4}{15}
\text{P(1 blue and 1 yellow)}=\dfrac{4}{15}+\dfrac{4}{15}
Marks = 2
Question 2
There are 5 red balls and 6 green balls in a bag.
One ball is drawn from the bag, then another without replacement.
2(a) Select the correct probability tree diagram to represent this information.
ANSWER: Multiple Choice (Type 1)
A:
B:
C:
D:
Answer: D
Workings:
Initially there are 5 red balls and 6 green balls in the bag, giving a total of 11, so the probabilities of selecting a red or green ball are \dfrac{5}{11} and \dfrac{6}{11} respectively.
Upon removal of one ball, there are a total of 10 balls remaining. If the first ball was red, the probability of selecting a second red is then \dfrac{4}{10}. If the first ball was green, the probability of selecting a second green is then \dfrac{5}{10}. The rest of the tree diagram can then be filled in on the basis that the probabilities for each event must add up to 1.
Marks = 3
2(b) Calculate the probability that one red and one green ball are taken from the bag.
Give your answer as a fraction in its simplest form.
ANSWER: Fraction
Answer: \dfrac{6}{11}
Workings:
\text{P(Red then Green)}= \dfrac{5}{11}\times\dfrac{6}{10}\dfrac{30}{110}
\text{P(Green then Red)}= \dfrac{6}{11}\times\dfrac{5}{10}=\dfrac{30}{110}
\text{P(1 Red and 1 Green)}=\dfrac{30}{110}+\dfrac{30}{110}=\dfrac{60}{110}=\dfrac{6}{11}
Marks = 2
2(c) What is the probability that the two balls drawn are the same colour?
Give your answer as a fraction in its simplest form.
ANSWER: Fraction
Answer: \dfrac{5}{11}
Workings:
There are two ways of calculating this.
Method 1:
The balls drawn can be either the same colour or different colours. Using the answer from the previous question,
\text{P(different colours)}=\dfrac{6}{11}
Hence, \text{P(same colour)}=1-\dfrac{6}{11}=\dfrac{5}{11}
Method 2:
\text{P(Red then Red)}=\dfrac{5}{11}\times\dfrac{4}{10}=\dfrac{20}{110}
\text{P(Green then Green)}=\dfrac{6}{11}\times\dfrac{5}{10}=\dfrac{30}{110}
\text{P(same colour)}=\dfrac{20}{110}+\dfrac{30}{110}=\dfrac{50}{110}=\dfrac{5}{11}
Marks = 2
Question 3
There are x balls in a bag.
8 of the balls are blue.
3 of the balls are green.
The rest of the balls are orange and pink.
Jake takes two balls from the bag without replacement.
The probability that he takes a blue then green ball is \dfrac{1}{10}.
Find the total number of balls in the bag.
ANSWER: Simple Text Answer
Answer: 16
Workings:
Probability of blue ball first: 8 blue balls and x balls in total, so
\text{P(Blue first)}=\dfrac{8}{x}
Probability of choosing a green ball after a ball isn’t replaced: 3 green balls and x-1 balls left in the bag, so
\text{P(Green second)}=\dfrac{3}{x-1}
Probabilities are independent, so multiply to find the probability of both happening, which is equal to \dfrac{1}{10}, and solve for x.
\dfrac{8}{x}\times\dfrac{3}{x-1}=\dfrac{1}{10}
\dfrac{24}{x(x-1)}=\dfrac{1}{10}
240=x(x-1)
240=x^2-x
x^2-x-240=0
(x-16)(x+15)=0
x=16 or x=-15
Since there cannot be a negative number of balls in the bag, there are 16 balls in total.
Marks = 5
Question 4
There are 10 counters in a bag.
3 are green and the rest are blue.
One counter is taken from the bag.
If the counter is green it is replaced, if the counter is blue it is not replaced.
A second counter is then taken from the bag.
Calculate the probability that two different counters are removed from the bag.
Give your answer as a fraction in its simplest form.
ANSWER: Fraction
\dfrac{133}{300}Workings:
\dfrac{21}{100} + \dfrac{21}{90} = \dfrac{399}{900} = \dfrac{133}{300}
Marks = 2
Question 5
A bag contains n counters
2 counters are and green the rest are black.
2 counters are taken at random without replacement.
5(a) Select from the list below the probability of selecting 2 green counters.
ANSWER: Multiple Choice (Type 1)
Answer: \dfrac{2}{n^2 - n}
\dfrac{4}{n}
0.25
\dfrac{2}{n^2}
Workings:
\dfrac{2}{n} \times \dfrac{1}{n-1} = \dfrac{2}{n(n-1)} = \dfrac{2}{n^2 - n}
Marks = 2
5(b) The probability that selecting both green counters is equal to 0.1
Calculate the smallest possible number of counters in the bag.
ANSWER: b
Workings
\frac{2}{n^2 -n} =0.1
20 = n^2 -n
n^2 -n-20 = 0
Using the quadratic formula
x=-4 and x=5
x=5 is the only real solution.
Marks = 2
Question 6
There are n sweets in a bag.
x of the sweets are lemon, the rest are lime.
Two sweets are taken at random from the bag without replacement.
A tree diagram to represent this information is shown below.
A, B and C represent probabilities.
5(a) Select the probability represented by A
ANSWER: Multiple Choice
Answer: \dfrac{n-x}{n}
\dfrac{x}{n-x}
\dfrac{x-1}{n}
\dfrac{n}{x}
Workings:
n is the total amount of sweets, lime represents the total amount (n) minus lemon (x).
\dfrac{n-x}{n}
Marks = 2
5(b) Select the probability represented by B
ANSWER: Multiple Choice
Answer: \dfrac{n-x}{n-1}
\dfrac{x}{n-1}
\dfrac{n-x}{n-x}
\dfrac{x}{n}
Workings
1 Sweet has been removed, which isn’t lime. The number of Lime sweets remains the same with 1 less from the total.
Marks = 2
5(c) Select the probability represented by C
ANSWER: Multiple Choice
Answer: \dfrac{n-x-1}{n-1}
\dfrac{n-x}{n}
\dfrac{n-x-1}{n}
\dfrac{n-x}{n-1}
Workings
1 Sweet has been removed, which is lime. The number of lime sweets goes down by 1 along with the total.
Marks = 2