Probability with Counting

When there are too many outcomes to list, you can still find probabilities: just count them instead. Permutations and combinations let you count both the outcomes you want and all the possible outcomes, and dividing gives the probability. This is how you find the chance of a particular poker hand or a lucky committee draw.

Key ideas

The basic formula, with counting

When all outcomes are equally likely (from sample spaces):

$$P(A) = \frac{n(A)}{n(S)} = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$

Now $n(A)$ and $n(S)$ are found with the counting principles, permutations, or combinations, instead of a list.

Count the top and bottom the same way

Decide once whether order matters, and use the same method for the numerator and the denominator. If you count the total with combinations, count the favourable outcomes with combinations too. (Using permutations for both also works, as long as you’re consistent, because the $r!$ cancels.)

A standard deck of cards

A standard deck has $52$ cards: $4$ suits (hearts and diamonds are red; clubs and spades are black), each with $13$ ranks (A, 2 to 10, J, Q, K). There are $4$ cards of each rank and $12$ face cards (J, Q, K). A $5$-card hand is a combination, so the number of possible hands is

$$\binom{52}{5} = 2\,598\,960$$

“At least one”: use the complement

The complement of “at least one” is “none”. Since $P(A) + P(A') = 1$:

$$P(\text{at least one}) = 1 - P(\text{none})$$

“None” is usually a single count, while “at least one” would need many cases.

Worked examples

Example 1: A random committee

A committee of $4$ is chosen at random from $6$ boys and $5$ girls. Find the probability that it has exactly $2$ girls.

Solution. Order doesn’t matter, so use combinations. Total committees: $\dbinom{11}{4} = 330$.

Favourable committees: $2$ of the $5$ girls and $2$ of the $6$ boys:

$$\binom{5}{2} \times \binom{6}{2} = 10 \times 15 = 150$$ $$P(\text{exactly 2 girls}) = \frac{150}{330} = \frac{5}{11}$$

Example 2: Card hands

A $5$-card hand is dealt from a well-shuffled standard deck. Find the probability that

• (a) all five cards are hearts
• (b) the hand has exactly $2$ aces

Solution. The total is $\dbinom{52}{5} = 2\,598\,960$ in both parts.

(a) Choose $5$ of the $13$ hearts: $\dbinom{13}{5} = 1287$.

$$P(\text{all hearts}) = \frac{1287}{2\,598\,960} = \frac{33}{66\,640} \approx 0.000\,495$$

(b) Choose $2$ of the $4$ aces and $3$ of the $48$ non-aces:

$$P(\text{exactly 2 aces}) = \frac{\binom{4}{2}\binom{48}{3}}{\binom{52}{5}} = \frac{6 \times 17\,296}{2\,598\,960} = \frac{103\,776}{2\,598\,960} \approx 0.0399$$

Example 3: Random arrangements

The letters of SQUARE are arranged at random. Find the probability that

• (a) the arrangement begins with a vowel
• (b) Q and U are next to each other

Solution. SQUARE has $6$ different letters, so there are $6! = 720$ equally likely arrangements.

(a) There are $3$ vowels (U, A, E) for the first spot, then $5!$ ways to arrange the rest:

$$P(\text{vowel first}) = \frac{3 \times 5!}{6!} = \frac{360}{720} = \frac{1}{2}$$

(b) Treat QU as a block: $5!$ arrangements, times $2!$ orders inside the block:

$$P(\text{Q and U together}) = \frac{5! \times 2!}{6!} = \frac{240}{720} = \frac{1}{3}$$

Example 4: At least one ace

A $5$-card hand is dealt. Find the probability that it contains at least one ace.

Solution. Use the complement. “No aces” means all $5$ cards come from the $48$ non-aces:

$$P(\text{no aces}) = \frac{\binom{48}{5}}{\binom{52}{5}} = \frac{1\,712\,304}{2\,598\,960} \approx 0.659$$ $$P(\text{at least one ace}) = 1 - \frac{1\,712\,304}{2\,598\,960} \approx 0.341$$

Common mistakes

Counting the top and bottom in different ways. Using $P(n, r)$ for the total but $\dbinom{n}{r}$ for the favourable outcomes gives a wrong answer. Pick one approach and stick with it.

Forgetting the “other” items. For “exactly $2$ aces in $5$ cards”, you also have to choose the other $3$ cards from the $48$ non-aces. Choosing them from all $50$ remaining cards would allow more aces.

Doing “at least one” the long way, or the wrong way. Use $1 - P(\text{none})$. Don’t multiply “one ace” by “any $4$ other cards”: that counts hands with two or more aces several times.

Rounding too early. Keep exact counts or fractions until the last step. Card probabilities are often tiny, so round to at least three significant digits.

A probability greater than $1$. If $n(A)$ is bigger than $n(S)$, you’ve counted the favourable outcomes in a way that allows repeats or order when the total doesn’t.

Practice

1. (Warm-up) Three students from a class of $10$ are chosen at random to win prizes (all the same). Kai is in the class. What is the probability that Kai wins?

Solution

Total: $\dbinom{10}{3} = 120$. With Kai included, choose the other $2$ from $9$: $\dbinom{9}{2} = 36$.

$$P(\text{Kai wins}) = \frac{36}{120} = \frac{3}{10}$$

(This makes sense: $3$ of the $10$ students win.)

2. (Warm-up) Four cards labelled A, B, C, D are shuffled and laid in a row. What is the probability they’re in the order ABCD?

Solution

There are $4! = 24$ equally likely orders, and only one is ABCD: $P = \dfrac{1}{24}$.

3. (Warm-up) Two cards are dealt from a standard deck. Find the probability that both are hearts.

Solution

$$\frac{\binom{13}{2}}{\binom{52}{2}} = \frac{78}{1326} = \frac{1}{17}$$

4. (Core) A $5$-card hand is dealt. Find the probability that all five cards are face cards (J, Q, or K).

Solution

There are $12$ face cards:

$$\frac{\binom{12}{5}}{\binom{52}{5}} = \frac{792}{2\,598\,960} \approx 0.000\,305$$

5. (Core) A committee of $3$ is chosen at random from $4$ teachers and $6$ students. Find the probability that it includes at least one teacher.

Solution

The complement is “no teachers”, meaning $3$ students:

$$P(\text{no teachers}) = \frac{\binom{6}{3}}{\binom{10}{3}} = \frac{20}{120} = \frac{1}{6}$$$$P(\text{at least one teacher}) = 1 - \frac{1}{6} = \frac{5}{6}$$

6. (Core) Seven people, including Zoe and Ali, line up in a random order. Find the probability that Zoe and Ali are next to each other.

Solution

Total: $7!$. Together: treat Zoe and Ali as a block, giving $6! \times 2!$.

$$P = \frac{6! \times 2!}{7!} = \frac{2}{7}$$

7. (Core) A $4$-digit code is made of $4$ different digits, chosen at random. Find the probability that all four digits are even.

Solution

Order matters for a code. Total: $P(10, 4) = 5040$. All even: arrange $4$ of the $5$ even digits ($0, 2, 4, 6, 8$): $P(5, 4) = 120$.

$$P = \frac{120}{5040} = \frac{1}{42}$$

(Using combinations for both gives the same answer: $\dbinom{5}{4} \div \dbinom{10}{4} = \dfrac{5}{210} = \dfrac{1}{42}$.)

8. (Challenge) A full house is a $5$-card hand with three cards of one rank and two of another (like three 8s and two kings). Find the probability of being dealt a full house.

Solution

Build the hand in stages and multiply:

• the rank for the three: $13$ ways; which $3$ of its $4$ suits: $\dbinom{4}{3} = 4$ ways
• the rank for the pair (a different rank): $12$ ways; which $2$ of its $4$ suits: $\dbinom{4}{2} = 6$ ways

$$13 \times 4 \times 12 \times 6 = 3744$$$$P(\text{full house}) = \frac{3744}{2\,598\,960} = \frac{6}{4165} \approx 0.001\,44$$

9. (Challenge) Five friends each pick a whole number from $1$ to $10$ at random. Find the probability that at least two of them pick the same number.

Solution

Order matters here (it matters who picked which number). Total: $10^5 = 100\,000$ equally likely ways. The complement is “all different”: $P(10, 5) = 10 \times 9 \times 8 \times 7 \times 6 = 30\,240$.

$$P(\text{at least two match}) = 1 - \frac{30\,240}{100\,000} = 0.6976$$

That’s almost $70\%$, much higher than most people guess!