Probability with Coins and Dice

One and Two Coins

One coin has sample space $\{H, T\}$ — two equally likely outcomes.

Two coins (or one coin tossed twice) has four equally likely outcomes: $$\{HH,\ HT,\ TH,\ TT\}.$$ From this list you can read off:

• $P(\text{two heads}) = \dfrac14$ (only $HH$),
• $P(\text{exactly one head}) = \dfrac24 = \dfrac12$ ($HT, TH$),
• $P(\text{at least one head}) = \dfrac34$ (all but $TT$),
• $P(\text{no head}) = \dfrac14$ (only $TT$).

Key Point: Write the whole sample space first; then counting favourable cases is easy.

One Die

One die has sample space $\{1,2,3,4,5,6\}$. Common events:

• $P(\text{even}) = P(\text{odd}) = \dfrac12$,
• $P(\text{prime}) = \dfrac36 = \dfrac12$ (primes $2,3,5$),
• $P(\text{multiple of 3}) = \dfrac26 = \dfrac13$ ($3,6$),
• $P(\text{number} > 4) = \dfrac26 = \dfrac13$ ($5,6$).

Key Point: For a single die the total is always 6 — only the favourable count changes.

Two Dice — 36 Outcomes

When two dice are thrown, each of the 36 ordered pairs $(a,b)$ with $a,b\in\{1,\dots,6\}$ is equally likely — total $= 6\times 6 = 36$.

The sum of the two dice ranges from 2 to 12, but the sums are not equally likely. Counting the pairs:

Sum	2	3	4	5	6	7	8	9	10	11	12
Ways	1	2	3	4	5	6	5	4	3	2	1

So $P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac16$ (the most likely sum), $P(\text{sum} = 2) = \dfrac1{36}$, and so on.

[Board Important] A doublet is a pair with equal numbers $\{(1,1),\dots,(6,6)\}$ — 6 of them, so $P(\text{doublet}) = \dfrac{6}{36} = \dfrac16$.

Reading the Two-Dice Grid

Think of the 36 outcomes as a $6\times 6$ grid. Any event is just a set of cells; its probability is (cells in the event) $\div$ 36. For example:

• $P(\text{both show 6}) = \dfrac1{36}$,
• $P(\text{sum} \le 4) = \dfrac{1+2+3}{36} = \dfrac{6}{36} = \dfrac16$,
• $P(\text{a total of 8}) = \dfrac{5}{36}$.

Key Point: Two dice $\Rightarrow$ 36 in the denominator. Count the favourable cells from the sum table.

Solved Examples

Example 1: Two coins

Two coins are tossed together. Find the probability of getting (i) at least one head, (ii) exactly one tail.

Solution:

1. Sample space $\{HH,HT,TH,TT\}$, total 4.
2. (i) At least one head $= \{HH,HT,TH\}$: $\dfrac34$.
3. (ii) Exactly one tail $= \{HT,TH\}$: $\dfrac24 = \dfrac12$.

Final Answer: (i) $\dfrac34$; (ii) $\dfrac12$.

Takeaway: List all 4 outcomes, then count.

Example 2: Sum of two dice

Two dice are thrown. Find the probability that the sum is (i) 7, (ii) 10, (iii) 8.

Solution:

1. Total outcomes $= 36$.
2. (i) Sum 7: $6$ ways $\Rightarrow \dfrac{6}{36} = \dfrac16$.
3. (ii) Sum 10: $\{(4,6),(5,5),(6,4)\}$, $3$ ways $\Rightarrow \dfrac{3}{36} = \dfrac1{12}$.
4. (iii) Sum 8: $5$ ways $\Rightarrow \dfrac{5}{36}$.

Final Answer: (i) $\dfrac16$; (ii) $\dfrac1{12}$; (iii) $\dfrac{5}{36}$.

Takeaway: Use the sum table for the number of ways.

Example 3: Doublets

Two dice are thrown. Find the probability of getting a doublet.

Solution:

1. Doublets $\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$: $6$ outcomes.
2. $P = \dfrac{6}{36} = \dfrac16$.

Final Answer: $\dfrac16$.

Takeaway: A doublet is "same number on both dice".

Example 4: Complement with dice

Two dice are thrown. Find the probability that the sum is not 7.

Solution:

1. $P(\text{sum} = 7) = \dfrac16$.
2. $P(\text{sum} \ne 7) = 1 - \dfrac16 = \dfrac56$.

Final Answer: $\dfrac56$.

Takeaway: Complement saves counting all the non-7 sums.