Comprehensive assessment covering all 10 lessons: set notation and Venn diagrams, the addition and complement rules, conditional probability, independent and mutually exclusive events, probability trees, discrete probability distributions, expected value and variance, measures of centre and spread, data displays, and their applications.
Assessment
Select the best answer for each question. 1 mark each.
If $P(A) = 0.35$, then the probability of the complement $P(A')$ is:
In a Venn diagram, $n(A) = 18$, $n(B) = 12$ and $n(A \cap B) = 5$. The value of $n(A \cup B)$ is:
Events $A$ and $B$ are mutually exclusive with $P(A) = 0.3$ and $P(B) = 0.45$. Then $P(A \cup B)$ equals:
Events $A$ and $B$ are independent with $P(A) = 0.6$ and $P(B) = 0.5$. Then $P(A \cap B)$ equals:
If $P(A \cap B) = 0.24$ and $P(B) = 0.6$, then the conditional probability $P(A \mid B)$ is:
Two fair six-sided dice are rolled. The probability that the sum of the two numbers is $7$ is:
A fair coin is tossed three times. Using a probability tree, the probability of getting at least one head is:
A bag holds $5$ red and $3$ blue counters. Two counters are drawn without replacement. The probability that both are red is:
Two events each have positive probability and are mutually exclusive. It follows that the two events are:
A discrete random variable $X$ takes the values $1, 2, 3, 4$ with probabilities $0.1, 0.2, 0.3, 0.4$ respectively. The expected value $E(X)$ is:
For a random variable $X$, $E(X) = 3$ and $E(X^2) = 13$. The variance $\text{Var}(X)$ is:
The median of the data set $4, 7, 9, 12, 15, 18$ is:
The interquartile range of the data set $3, 5, 7, 9, 11, 13, 15, 17$ is:
A data set has standard deviation $4$. If every value is multiplied by $3$, the new standard deviation is:
A data set is positively skewed (right-skewed). Which statement is most likely true?
Short Answer
In a group of $30$ students, $18$ study History, $14$ study Geography, and $6$ study both. Using a Venn diagram, find (a) the number who study only History, and (b) the number who study neither subject. Show all working.
For two events, $P(A) = 0.5$, $P(B) = 0.4$ and $P(A \cup B) = 0.7$. Find (a) $P(A \cap B)$, and (b) determine whether $A$ and $B$ are independent, justifying your answer. Show all working.
A box contains $4$ green and $6$ yellow balls. Two balls are drawn without replacement. Using a probability tree, find (a) $P(\text{both yellow})$, and (b) $P(\text{one of each colour})$. Show all working.
The two-way table below records $200$ people by whether they exercise and whether they are aged over $40$.
| Exercises | Does not | Total | |
|---|---|---|---|
| Over 40 | 45 | 35 | 80 |
| 40 or under | 90 | 30 | 120 |
| Total | 135 | 65 | 200 |
One person is chosen at random. Find (a) $P(\text{exercises})$, and (b) $P(\text{over 40} \mid \text{exercises})$. Show all working.
A discrete random variable $X$ has the probability distribution below.
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(X=x)$ | 0.2 | 0.3 | $k$ | 0.1 |
Find (a) the value of $k$, and (b) the expected value $E(X)$. Show all working.
For the data set $12, 15, 15, 18, 20, 22, 28$, find (a) the median, (b) the lower quartile $Q_1$ and upper quartile $Q_3$, and (c) the interquartile range. Show all working.
The heights (cm) of five plants are $20, 24, 26, 28, 32$. Find (a) the mean, and (b) the population standard deviation, correct to two decimal places. Show all working.
A factory has two machines. Machine A makes $60\%$ of the items and Machine B makes $40\%$. Machine A produces $5\%$ defective items, while Machine B produces $8\%$ defective. An item is selected at random. Using a probability tree, find (a) the probability the item is defective, and (b) given that it is defective, the probability it came from Machine A. Give answers to three decimal places. Show all working.
Q1: B, complement rule: $P(A') = 1 - P(A) = 1 - 0.35 = 0.65$.
Q2: A, inclusionâexclusion: $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 18 + 12 - 5 = 25$.
Q3: C, mutually exclusive means $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B) = 0.3 + 0.45 = 0.75$.
Q4: B, for independent events $P(A \cap B) = P(A) \times P(B) = 0.6 \times 0.5 = 0.3$.
Q5: C, $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.24}{0.6} = 0.4$.
Q6: A, six of the $36$ outcomes sum to $7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, so $P = \frac{6}{36} = \frac{1}{6}$.
Q7: D, use the complement: $P(\text{no heads}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$, so $P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8}$.
Q8: A, $P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$.
Q9: B, if $P(A) > 0$ and $P(B) > 0$ but $P(A \cap B) = 0$, then $P(A) \times P(B) \neq 0 = P(A \cap B)$, so they cannot be independent, they are dependent.
Q10: B, $E(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0$.
Q11: A, $\text{Var}(X) = E(X^2) - [E(X)]^2 = 13 - 3^2 = 13 - 9 = 4$.
Q12: B, with $6$ values the median is the mean of the $3$rd and $4$th values: $\frac{9 + 12}{2} = 10.5$.
Q13: D, lower half $3,5,7,9$ gives $Q_1 = \frac{5+7}{2} = 6$; upper half $11,13,15,17$ gives $Q_3 = \frac{13+15}{2} = 14$; so $\text{IQR} = 14 - 6 = 8$.
Q14: C, multiplying every value by $3$ multiplies the standard deviation by $3$: $4 \times 3 = 12$.
Q15: D, a positive (right) skew has a long upper tail that pulls the mean above the median, so mean $>$ median.
Q16 (3 marks): Place $6$ in the intersection (study both) [1]. History only $= 18 - 6 = 12$, Geography only $= 14 - 6 = 8$, so $n(\text{History or Geography}) = 12 + 6 + 8 = 26$ [1]. (a) History only $= 12$. (b) Neither $= 30 - 26 = 4$ students [1].
Q17 (3 marks): (a) $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, so $0.7 = 0.5 + 0.4 - P(A \cap B)$, giving $P(A \cap B) = 0.2$ [1]. (b) For independence check $P(A) \times P(B) = 0.5 \times 0.4 = 0.2$ [1]. Since $P(A) \times P(B) = 0.2 = P(A \cap B)$, the events are independent [1].
Q18 (3 marks): First draw: $P(\text{yellow}) = \frac{6}{10}$, $P(\text{green}) = \frac{4}{10}$ [1]. (a) $P(\text{both yellow}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$ [1]. (b) $P(\text{one of each}) = P(GY) + P(YG) = \frac{4}{10} \times \frac{6}{9} + \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}$ [1].
Q19 (3 marks): (a) $P(\text{exercises}) = \frac{135}{200} = 0.675$ [1]. (b) Restrict to the $135$ who exercise; of these, $45$ are over $40$ [1], so $P(\text{over 40} \mid \text{exercises}) = \frac{45}{135} = \frac{1}{3} \approx 0.333$ [1].
Q20 (4 marks): (a) Probabilities sum to $1$: $0.2 + 0.3 + k + 0.1 = 1$ [1], so $k = 1 - 0.6 = 0.4$ [1]. (b) $E(X) = 0(0.2) + 1(0.3) + 2(0.4) + 3(0.1)$ [1] $= 0 + 0.3 + 0.8 + 0.3 = 1.4$ [1].
Q21 (4 marks): Data ordered: $12, 15, 15, 18, 20, 22, 28$ with $n = 7$. (a) Median is the $4$th value $= 18$ [1]. (b) Lower half $12, 15, 15$ gives $Q_1 = 15$ [1]; upper half $20, 22, 28$ gives $Q_3 = 22$ [1]. (c) $\text{IQR} = Q_3 - Q_1 = 22 - 15 = 7$ [1].
Q22 (4 marks): (a) Mean $\bar{x} = \frac{20 + 24 + 26 + 28 + 32}{5} = \frac{130}{5} = 26$ [1]. (b) Deviations from the mean: $-6, -2, 0, 2, 6$; squared: $36, 4, 0, 4, 36$, which sum to $80$ [1]. Variance $= \frac{80}{5} = 16$ [1]. Population standard deviation $= \sqrt{16} = 4.00$ cm [1].
Q23 (4 marks): Tree branches: $P(A) = 0.6$ with $P(\text{def} \mid A) = 0.05$; $P(B) = 0.4$ with $P(\text{def} \mid B) = 0.08$. (a) $P(\text{defective}) = 0.6 \times 0.05 + 0.4 \times 0.08 = 0.03 + 0.032 = 0.062$ [1] [1]. (b) $P(A \mid \text{defective}) = \frac{P(A \cap \text{defective})}{P(\text{defective})} = \frac{0.03}{0.062}$ [1] $\approx 0.484$ [1].