Using counting in probability

Find probabilities by counting the sample space and the event space with permutations and combinations, using the complement and a mix of methods.

Counting both spaces

Example 1●●

A committee of $3$ is chosen at random from $5$ men and $4$ women. Find the probability that it has exactly $2$ women.

Sample space: ${}^9C_3 = 84$.

Event: ${}^4C_2 \times {}^5C_1 = 6 \times 5 = 30$.

$P = \dfrac{30}{84} = \dfrac{5}{14}$

Example 2●●

The letters of PARRAMATTA are arranged at random. Find the probability that it starts with P.

Sample space: $\dfrac{10!}{4!\,2!\,2!} = 37\,800$ (four As, two Rs, two Ts).

Event: $\dfrac{9!}{4!\,2!\,2!} = 3780$.

Fix P first, then arrange the other 9 letters.

$P = \dfrac{3780}{37\,800} = \dfrac{1}{10}$, which makes sense: P is one of the $10$ letters.

Example 3●●●

Six people, including Ali and Bea, line up at random. Find the probability that Ali and Bea are not next to each other.

Sample space: $6! = 720$.

Together: $5! \times 2 = 240$, so $P(\text{together}) = \dfrac{240}{720} = \dfrac13$.

$P(\text{not together}) = 1 - \dfrac13 = \dfrac23$

Example 4●●●

Five cards are dealt from a standard pack. Find the probability of at least one ace, to 3 decimal places.

No aces: $\dfrac{{}^{48}C_5}{{}^{52}C_5} = \dfrac{1\,712\,304}{2\,598\,960} \approx 0.659$

At least one ace: $1 - 0.659 \approx 0.341$

Flashcards

$P(E)$ by counting

$\dfrac{|E|}{|S|}$, with both counted the same way.

At least one

$1 - P(\text{none})$

Exactly $k$ women in a committee

${}^wC_k \times {}^mC_{r-k}$ over ${}^{w+m}C_r$

Practice

How many committees of $3$ can be chosen from $9$ people?
A committee of $3$ from $5$ men and $4$ women. Find the probability of exactly $2$ women.
Find the probability that the committee is all women.
The letters of PARRAMATTA are arranged at random. Find the probability that it starts with P.
Six people line up at random. Find the probability that Ali and Bea are not next to each other.
Five cards are dealt. Find the probability of at least one ace, to 3 decimal places.
Your event count is larger than your sample space count. The most likely cause is:
Three people sit in a row at random. Find the probability that a particular person sits in the middle.
Four letters are chosen from ABCDEFGH. The best way to count the sample space when only the set of letters matters is: