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If you roll a die AND flip a coin at the same time, how many different results are possible? How would you list them all?
Every possible result of an experiment, listed with no gaps and no repeats.
Students often think $(1,2)$ and $(2,1)$ are the same outcome when rolling two dice. They are not — ordered pairs mean the first die shows 1 and second shows 2 in the first case, and vice versa. Always list outcomes as ordered pairs. There are always exactly $6 \times 6 = 36$ outcomes for two dice — not 21.
The sample space $S$ lists every possible outcome. For equally likely outcomes:
$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total outcomes in }S}$$Example: Rolling a die — $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$ because the favourable outcomes are $\{2,\,4,\,6\}$.
For compound experiments, list outcomes as ordered pairs. Work through one variable at a time.
Die and coin: Fix the die face, list both coin outcomes, then move to next die face:
$(1,H),\;(1,T),\;(2,H),\;(2,T),\;(3,H),\;(3,T),$
$(4,H),\;(4,T),\;(5,H),\;(5,T),\;(6,H),\;(6,T)$
Total: $6 \times 2 = 12$ outcomes. $P(\text{odd and H}) = \dfrac{3}{12} = \dfrac{1}{4}$ (outcomes: $(1,H),(3,H),(5,H)$).
If experiment A has $m$ outcomes and B has $n$ outcomes, the combined experiment has $m \times n$ outcomes. This extends to any number of stages.
Worked example — three coins:
$P(\text{all heads}) = \dfrac{1}{8}$ and $P(\text{exactly two heads}) = \dfrac{3}{8}$ (HHT, HTH, THH).
Rolling two dice: 36 ordered pairs. To find $P(\text{sum} > 8)$, list all pairs where $d_1 + d_2 > 8$:
$(3,6),\;(4,5),\;(4,6),\;(5,4),\;(5,5),\;(5,6),\;(6,3),\;(6,4),\;(6,5),\;(6,6)$
That is 10 favourable outcomes.
$$P(\text{sum} > 8) = \frac{10}{36} = \frac{5}{18}$$Always verify: number of favourable outcomes $\leq 36$. ✓
Sample space ($S$): the complete set of all possible outcomes.
Counting principle: $|S| = m \times n$ for a two-stage experiment with $m$ and $n$ outcomes.
$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total outcomes in }S}$$For two dice: $|S| = 36$. List ordered pairs $(d_1, d_2)$ — order matters.
When rolling two standard dice, how many outcomes are in the sample space?
A coin is flipped and a 4-sector spinner is spun. How many outcomes are in the sample space?
Two fair coins are flipped. What is $P(HH)$?
Two dice are rolled. What is $P(\text{sum} = 7)$?
Which of the following is the correct sample space for rolling a single standard die?
Q6. A coin is flipped three times. (a) List the full sample space — all 8 outcomes. (b) Find $P(\text{exactly 2 heads})$.
Q7. A bag contains one Red (R), one Blue (B), and one Green (G) marble. A marble is drawn and a die is also rolled. (a) How many outcomes are in the sample space? (b) Find $P(\text{Red marble and even number})$.
Q8. Two dice are rolled. Find $P(\text{sum} \leq 4)$ by listing all outcomes where the sum is 4 or less. Show all working.
Sample space: $\{HHH,\,HHT,\,HTH,\,HTT,\,THH,\,THT,\,TTH,\,TTT\}$ — 8 outcomes.
Exactly 2 heads: $HHT,\,HTH,\,THH$ — 3 favourable outcomes.
$P(\text{exactly 2 heads}) = \dfrac{3}{8}$
Counting principle: $3 \times 6 = 18$ outcomes in sample space.
Favourable (Red AND even): $(R,2),\,(R,4),\,(R,6)$ — 3 outcomes.
$P(\text{R and even}) = \dfrac{3}{18} = \dfrac{1}{6}$
Outcomes with sum $\leq 4$: $(1,1),\,(1,2),\,(1,3),\,(2,1),\,(2,2),\,(3,1)$ — 6 favourable outcomes.
$P(\text{sum} \leq 4) = \dfrac{6}{36} = \dfrac{1}{6}$
A spinner has the numbers 1, 2, 3 and another spinner has the letters A, B, C, D. One result from each spinner is chosen. (a) List all 12 outcomes in the sample space. (b) Find $P(\text{number is prime AND letter comes before C alphabetically})$. Show your working clearly.