Covers Lessons 1–5: sample space and events, probability rules, conditional probability, independence and mutual exclusivity, and discrete probability distributions.
Assessment
Select the best answer for each question.
A bag contains 5 red, 3 blue, and 2 green marbles. One marble is drawn at random. What is $P(\text{not blue})$?
If $P(A) = 0.4$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$, what is $P(A \cup B)$?
Two events $A$ and $B$ are independent with $P(A) = 0.3$ and $P(B) = 0.4$. What is $P(A \cap B)$?
Given $P(A \cap B) = 0.15$ and $P(B) = 0.5$, what is $P(A \mid B)$?
Two events are mutually exclusive with $P(A) = 0.2$ and $P(B) = 0.5$. Which statement is true?
A random variable $X$ has $E(X) = 4$ and $E(X^2) = 20$. What is $\text{Var}(X)$?
A fair coin is flipped three times. What is $P(\text{at least one head})$?
If $Y = 3X + 2$ where $\text{Var}(X) = 4$, what is $\text{Var}(Y)$?
Short Answer
In a class of 40 students, 22 study Biology, 18 study Chemistry, and 10 study both. A student is selected at random. (a) Find $P(\text{Biology} \cup \text{Chemistry})$. (b) Find $P(\text{Chemistry} \mid \text{Biology})$. (c) Determine whether studying Biology and studying Chemistry are independent events, showing all working.
A factory has two machines. Machine A produces 60% of items with a 3% defect rate. Machine B produces 40% of items with a 5% defect rate. (a) Draw a tree diagram showing this information. (b) Find the probability that a randomly selected item is defective. (c) Given that an item is defective, find the probability it came from Machine A.
A random variable $X$ has the following probability distribution:
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $P(X=x)$ | $0.2$ | $0.3$ | $0.3$ | $0.2$ |
(a) Verify that this is a valid probability distribution. (b) Find $E(X)$. (c) Find $\text{Var}(X)$.
Q1: C — $P(\text{not blue}) = 1 - \frac{3}{10} = \frac{7}{10}$.
Q2: C — $P(A \cup B) = 0.4 + 0.5 - 0.2 = 0.7$.
Q3: C — $P(A \cap B) = 0.3 \times 0.4 = 0.12$ (independent).
Q4: B — $P(A \mid B) = \frac{0.15}{0.5} = 0.3$.
Q5: B — Mutually exclusive events with positive probability are dependent. $P(A \cap B) = 0 \neq 0.2 \times 0.5$.
Q6: B — $\text{Var}(X) = 20 - 16 = 4$.
Q7: B — $P(\text{at least one H}) = 1 - (\frac{1}{2})^3 = 1 - \frac{1}{8} = \frac{7}{8}$.
Q8: C — $\text{Var}(Y) = 3^2 \times 4 = 36$.
Q9 (3 marks): (a) $P(B \cup C) = \frac{22}{40} + \frac{18}{40} - \frac{10}{40} = \frac{30}{40} = \frac{3}{4} = 0.75$ [1]. (b) $P(C \mid B) = \frac{P(B \cap C)}{P(B)} = \frac{10/40}{22/40} = \frac{10}{22} = \frac{5}{11}$ [1]. (c) $P(B) \times P(C) = \frac{22}{40} \times \frac{18}{40} = \frac{396}{1600} = 0.2475$. $P(B \cap C) = \frac{10}{40} = 0.25$. Since $0.25 \neq 0.2475$, the events are not independent (they are dependent) [1].
Q10 (3 marks): (a) Tree: A (0.6) → Def (0.03), OK (0.97); B (0.4) → Def (0.05), OK (0.95) [0.5]. (b) $P(\text{Def}) = 0.6 \times 0.03 + 0.4 \times 0.05 = 0.018 + 0.020 = 0.038$ [1.5]. (c) $P(A \mid \text{Def}) = \frac{0.6 \times 0.03}{0.038} = \frac{0.018}{0.038} = \frac{18}{38} = \frac{9}{19} \approx 0.474$ [1].
Q11 (3 marks): (a) All probabilities are between 0 and 1. Sum = $0.2 + 0.3 + 0.3 + 0.2 = 1.0$ ✓ [0.5]. (b) $E(X) = 0(0.2) + 1(0.3) + 2(0.3) + 3(0.2) = 0 + 0.3 + 0.6 + 0.6 = 1.5$ [1]. (c) $E(X^2) = 0(0.2) + 1(0.3) + 4(0.3) + 9(0.2) = 0 + 0.3 + 1.2 + 1.8 = 3.3$ [0.5]. $\text{Var}(X) = 3.3 - 1.5^2 = 3.3 - 2.25 = 1.05$ [1].