MICAT Linear Programming — practice questions

1. Your syllabus: the sample space $S$ is the set of ALL possible outcomes of a random experiment. 2. A six-faced die can land on faces showing $1, 2, 3, 4, 5,$ or $6$. 3. So $S = \{1, 2, 3, 4, 5, 6\}$. 4. Option B starts at 0 — not how dice work. Option C is the EVEN-face subset; Option D is the ODD-face subset. Neither is the full sample space. _Source: syllabus Mathematics, Ch 14 "Probability", §14.1 (Sample space), p. 1._

1. For a fair coin, the sample space is $S = \{H, T\}$ with each outcome equally likely. 2. Probability of an event $E$ for equally-likely outcomes: $P(E) = \dfrac{|E|}{|S|}$. 3. Here $E = \{H\}$, $|E| = 1$, $|S| = 2$. So $P(E) = 1/2$. 4. Probabilities always lie in $[0, 1]$: option A (0) means impossible event; option C (1) means certain event. 5. Option D ($1/4$) would correspond to TWO coin tosses both being heads, not one. _Source: syllabus Mathematics, Ch 14, §14.2.2 (Probability of equally likely outcomes), p. 7._

1. Sample space: $S = \{1, 2, 3, 4, 5, 6\}$, with 6 equally likely outcomes. 2. Event 'odd number': $E = \{1, 3, 5\}$, with $|E| = 3$ outcomes. 3. $P(E) = 3/6 = 1/2$. 4. Equivalently, half the outcomes are odd → probability is $1/2$, an intuitive answer. 5. Option A is for ONE specific number (like $P(\text{rolling 5}) = 1/6$). Option B is for two specific values out of six. **Option D is the probability of NOT being odd (i.e., being even), which is the same $1/2$ — but we're asked for odd, not even.** _Source: syllabus Mathematics, Ch 14, §14.2.2 (Equally likely outcomes — die examples), p. 7–8._