16.5 Probability Using Counting Principles
AHL 1.10
- how to find probabilities by counting with combinations \({}^n\mathrm{C}_r\) and permutations
- when to use combinations (order does not matter) and when to use arrangements (order matters)
- how to combine counting with the complement rule, conditional probability and the test for independence
Counting the outcomes
When all the outcomes are equally likely, \(\mathrm{P}(A) = \dfrac{n(A)}{n(U)}\) (Section 16.1). If the numbers \(n(A)\) and \(n(U)\) are too large to list, we count them with the methods of Unit 5.
Probability by counting
\[ \mathrm{P}(A) = \frac{\text{number of selections (or arrangements) in } A}{\text{total number of selections (or arrangements)}} \]
- Selections (a group, a hand of cards, a handful of counters; order does not matter): use \({}^n\mathrm{C}_r = \dbinom{n}{r}\).
- Arrangements (a row, a word, a number; order matters): use \(n!\) and \({}^n\mathrm{P}_r\).
- Count the top and the bottom in the same way.
- \(A\): the event
- \(n\): the number of objects to choose from
- \(r\): the number of objects chosen
- \({}^n\mathrm{C}_r = \dbinom{n}{r}\): the number of selections of \(r\) objects from \(n\)
- \(n!\): \(n\) factorial, the number of arrangements of \(n\) objects in a row
- \({}^n\mathrm{P}_r\): the number of arrangements of \(r\) objects from \(n\)
\({}^n\mathrm{C}_r = \dfrac{n!}{r!\,(n - r)!}\) is in the formula booklet; \({}^n\mathrm{P}_r = \dfrac{n!}{(n - r)!}\) is in the HL formula booklet.
Choosing \(r\) objects “at random without replacement” is the same as choosing a random group of \(r\), so we can count groups with \({}^n\mathrm{C}_r\) instead of drawing a tree diagram.
Example 1 A bag contains 5 blue counters and 7 yellow counters. Four counters are taken at random, without replacement. Find the probability that
(a) all four counters are yellow,
(b) at least one counter of each colour is taken,
(c) at least two yellow counters are taken,
(d) at least two yellow counters are taken, given that at least one counter of each colour is taken.
(e) State, with a reason, whether the events “at least two yellow” and “at least one of each colour” are independent.
Total number of groups of 4 from 12: \(\dbinom{12}{4} = 495\).
(a) \(\dbinom{7}{4} = 35\) groups are all yellow: \(\;\mathrm{P} = \dfrac{35}{495} = \dfrac{7}{99}\).
(b) Use the complement: not “all yellow” and not “all blue”. All blue: \(\dbinom{5}{4} = 5\).
\[ \mathrm{P} = 1 - \frac{35 + 5}{495} = \frac{455}{495} = \frac{91}{99} . \]
(c) Complement: 0 yellow (\(\dbinom{5}{4} = 5\) ways) or exactly 1 yellow (\(\dbinom{7}{1}\dbinom{5}{3} = 70\) ways).
\[ \mathrm{P} = 1 - \frac{5 + 70}{495} = \frac{420}{495} = \frac{28}{33} . \]
(d) “At least two yellow and at least one of each” means exactly 2 or exactly 3 yellow:
\[ \binom{7}{2}\binom{5}{2} + \binom{7}{3}\binom{5}{1} = 21 \times 10 + 35 \times 5 = 385 . \]
Given “at least one of each” (\(455\) groups), \(\;\mathrm{P} = \dfrac{385}{455} = \dfrac{11}{13}\).
(e) \(\mathrm{P}(\text{at least two yellow} \mid \text{one of each}) = \dfrac{11}{13} \approx 0.846\), but \(\mathrm{P}(\text{at least two yellow}) = \dfrac{28}{33} \approx 0.848\). They are not equal, so the events are not independent (although they are very close).
Example 2 The six letters of the word NUMBER are arranged at random in a row. Find the probability that
(a) the arrangement starts and ends with a vowel,
(b) N and M are next to each other,
(c) N and M are not next to each other.
There are \(6! = 720\) equally likely arrangements.
(a) The vowels are U and E. Ends first: \(2 \times 1 = 2\) ways. The middle 4 letters: \(4! = 24\).
\[ \mathrm{P} = \frac{2 \times 24}{720} = \frac{48}{720} = \frac{1}{15} . \]
(b) Glue N and M into one block: \(5!\) arrangements, and \(2\) orders inside the block.
\[ \mathrm{P} = \frac{5! \times 2}{6!} = \frac{240}{720} = \frac{1}{3} . \]
(c) \(1 - \dfrac{1}{3} = \dfrac{2}{3}\).
Example 3 A committee of 5 is chosen at random from 6 boys and 4 girls. Find the probability that the committee has
(a) exactly 2 girls, \(\qquad\) (b) at least one girl.
Total: \(\dbinom{10}{5} = 252\).
(a) \(\dbinom{4}{2}\dbinom{6}{3} = 6 \times 20 = 120\), so \(\;\mathrm{P} = \dfrac{120}{252} = \dfrac{10}{21}\).
(b) No girls: \(\dbinom{6}{5} = 6\). \(\;\mathrm{P} = 1 - \dfrac{6}{252} = \dfrac{246}{252} = \dfrac{41}{42}\).
Example 4 Five cards are dealt at random from an ordinary pack of 52 cards. Find the probability that the hand contains exactly two aces.
Choose 2 of the 4 aces, and 3 of the 48 other cards:
\[ \mathrm{P} = \frac{\dbinom{4}{2}\dbinom{48}{3}}{\dbinom{52}{5}} = \frac{6 \times 17\,296}{2\,598\,960} = \frac{103\,776}{2\,598\,960} \approx 0.0399 . \]
Calculator page → Menu → Probability → Combinations gives nCr(. For Example 4 type
nCr(4,2)*nCr(48,3)/nCr(52,5)
and press ctrl + enter for a decimal: \(0.039929\ldots \approx 0.0399\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Top counted with \({}^n\mathrm{P}_r\), bottom with \({}^n\mathrm{C}_r\) | Use the same method for both | Ordered and unordered counts cannot be mixed in one fraction. |
| “At least two yellow” \(= \binom{7}{2}\binom{10}{2}\) | Add separate cases, or use the complement | Choosing “2 yellow, then any 2” counts some groups more than once. |
| Glued block, forgetting the \(\times 2\) | \(5! \times 2\) | NM and MN are different arrangements. |
| Conditional probability with the full total \(495\) as the denominator | Use the given event’s count, \(455\) | Given an event, the sample space shrinks to that event. |
Exercises
1 Three students are chosen at random from 5 boys and 4 girls. Find the probability that
(a) all three are girls, \(\qquad\) (b) exactly two are boys.
Total \(\dbinom{9}{3} = 84\).
(a) \(\dfrac{\binom{4}{3}}{84} = \dfrac{4}{84} = \dfrac{1}{21}\).
(b) \(\dfrac{\binom{5}{2}\binom{4}{1}}{84} = \dfrac{40}{84} = \dfrac{10}{21}\).
2 The letters of the word SQUARE are arranged at random. Find the probability that
(a) the arrangement starts with S,
(b) the three vowels are all next to each other.
Total \(6! = 720\).
(a) \(\dfrac{5!}{6!} = \dfrac{1}{6}\).
(b) Glue U, A, E into one block: \(4!\) arrangements of the block and S, Q, R; \(3!\) inside the block. \(\;\dfrac{4! \times 3!}{720} = \dfrac{144}{720} = \dfrac{1}{5}\).
3 Six people, including Ali and Bo, stand in a row in a random order. Find the probability that Ali and Bo are at the two ends of the row.
Ali and Bo at the ends: \(2\) ways; the other 4 in the middle: \(4!\). \(\;\mathrm{P} = \dfrac{2 \times 24}{720} = \dfrac{1}{15}\).
4 In a lottery, 6 different numbers are chosen at random from \(1\) to \(40\). Find the probability that a ticket with 6 numbers matches all six.
\(\dfrac{1}{\binom{40}{6}} = \dfrac{1}{3\,838\,380}\).
5 Paper 2 Four cards are dealt from an ordinary pack of 52 cards. A pack has 12 picture cards (jacks, queens and kings). Find the probability that the hand contains no picture cards.
\(\dfrac{\binom{40}{4}}{\binom{52}{4}} = \dfrac{91\,390}{270\,725} \approx 0.338\).
6 The digits \(1, 2, 3, 4, 5, 6\) are arranged at random to make a six-digit number (no repeats). Find the probability that the number is
(a) even, \(\qquad\) (b) greater than \(500\,000\) and even.
Total \(6! = 720\).
(a) The last digit is \(2\), \(4\) or \(6\): \(\;\dfrac{3 \times 5!}{720} = \dfrac{1}{2}\).
(b) First digit 5: last digit 2, 4 or 6, so \(3 \times 4! = 72\). First digit 6: last digit 2 or 4, so \(2 \times 4! = 48\). \(\;\mathrm{P} = \dfrac{72 + 48}{720} = \dfrac{1}{6}\).
7 A bag holds 3 red, 4 green and 5 white counters. Four are taken at random without replacement. Find the probability that
(a) exactly two are white, \(\qquad\) (b) none is white.
Total \(\dbinom{12}{4} = 495\).
(a) \(\dfrac{\binom{5}{2}\binom{7}{2}}{495} = \dfrac{210}{495} = \dfrac{14}{33}\).
(b) \(\dfrac{\binom{7}{4}}{495} = \dfrac{35}{495} = \dfrac{7}{99}\).
8 Eight people, including three friends, sit in a row of eight seats in a random order. Find the probability that the three friends sit next to each other.
Glue the friends into one block: \(6!\) arrangements, \(3!\) inside the block. \(\;\mathrm{P} = \dfrac{6! \times 3!}{8!} = \dfrac{6}{56} = \dfrac{3}{28}\).
9 A team of 4 is chosen at random from 5 men and 5 women. Find the probability that the team has more women than men.
3 women and 1 man: \(\dbinom{5}{3}\dbinom{5}{1} = 50\). 4 women: \(\dbinom{5}{4} = 5\). Total \(\dbinom{10}{4} = 210\).
\(\mathrm{P} = \dfrac{55}{210} = \dfrac{11}{42}\).
10 A drawer contains 4 black socks and 6 white socks. Two socks are taken at random.
(a) Find the probability that the two socks are the same colour.
(b) Given that they are the same colour, find the probability that both are black.
Total \(\dbinom{10}{2} = 45\). Both black: \(\dbinom{4}{2} = 6\). Both white: \(\dbinom{6}{2} = 15\).
(a) \(\dfrac{6 + 15}{45} = \dfrac{21}{45} = \dfrac{7}{15}\).
(b) \(\dfrac{6}{21} = \dfrac{2}{7}\).