Quick Revision

The Basic Rule

An experiment (tossing a coin, drawing a card) has outcomes; the list of all of them is the sample space.
An event is a group of outcomes. When every outcome is equally likely, $P(E) = \frac{\text{favourable outcomes}}{\text{total outcomes}}$
$0 \le P(E) \le 1$ . A sure event has probability 1 and an impossible one 0.
Give answers in lowest terms; $\frac{3}{20}$ , 0.15 and 15% are the same probability.

For example: a number from 1 to 20 is a multiple of 6 in 3 ways (6, 12, 18), so $P = \frac{3}{20}$ .

Complement and Odds

$P(\text{not } E) = 1 - P(E)$ . Use it for 'not' and 'at least one'.
Odds in favour a : b mean $P(E) = \frac{a}{a + b}$ .
Odds against a : b mean $P(E) = \frac{b}{a + b}$ .
From $P(E) = \frac{p}{q}$ , the odds in favour are $p : (q - p)$ and the odds against are $(q - p) : p$ .

For example: odds in favour of 5 : 6 give $P(E) = \frac{5}{11}$ and $P(\text{not } E) = \frac{6}{11}$ ; the odds against are 6 : 5.

Numbers and Letters

Multiples of k from 1 to n: n divided by k, remainder dropped.
Multiples of a or b: add both counts, then subtract the multiples of the LCM.
Primes: 15 up to 50, 25 up to 100; 1 is not a prime.
Perfect squares: 7 up to 50, 10 up to 100. Two-digit numbers: 90.
A letter from a word: count repeats too. ACCOUNTANT: 10 letters, 4 vowels.

For example: from 1 to 60, multiples of 5 or 6 number $12 + 10 - 2 = 20$ , so $P = \frac{1}{3}$ .

Calendar Questions

An ordinary year is 52 weeks and 1 spare day; a leap year is 52 weeks and 2 spare days in a row.
A leap year's spare pair is one of 7: (Sun, Mon), (Mon, Tue), and so on to (Sat, Sun).
53 Sundays: $\frac{1}{7}$ in an ordinary year and $\frac{2}{7}$ in a leap year.
In a leap year, 53 of either of two neighbouring days: $\frac{3}{7}$ ; 53 of both: $\frac{1}{7}$ .
53 of either of two days that are not neighbours, such as Sunday or Thursday: $\frac{4}{7}$ .

Coins

n coins, or one coin tossed n times: $2^{n}$ outcomes (2, 4, 8, 16, 32).
Exactly k heads: $\frac{^{n}C_{k}}{2^{n}}$ . The counts run 1, 3, 3, 1 for three coins and 1, 4, 6, 4, 1 for four.
At least one head: $1 - \frac{1}{2^{n}}$ .
One head and one tail with two coins is HT or TH, so $\frac{1}{2}$ .
Tossing until the first head: it comes on toss k with probability $\frac{1}{2^{k}}$ .

For example: three coins, at most one head: $\frac{1 + 3}{8} = \frac{1}{2}$ .

One Die and Two Dice

One die: even 3, prime 3 (2, 3, 5), more than 4: 2.
Two dice: 36 outcomes; (2, 5) and (5, 2) are different.
A sum s has $6 - |s - 7|$ ways, so the sums 2 to 12 have 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 ways.
A coin and a die: 12 outcomes, or multiply the two chances.

For example: a sum of 5 or less has $1 + 2 + 3 + 4 = 10$ ways, so $P = \frac{10}{36} = \frac{5}{18}$ .

Two-Dice Counts (out of 36)

Doublet 6; sum even 18; sum prime 15; sum a multiple of 3: 12; sum a multiple of 4: 9.
Product odd 9 (both numbers odd); product even 27.
At least one six 11; exactly one six 10.
A difference of k has $2(6 - k)$ ways: 10, 8, 6, 4 and 2 for k = 1 to 5.
Three dice: 216 outcomes. All three the same: 6 ways, so $\frac{1}{36}$ . Sum 18: 1 way. Sum 17: 3 ways.

The Pack of 52

4 suits of 13: spades and clubs black, hearts and diamonds red.
13 ranks of 4 cards each: A, 2 to 10, J, Q, K.
Face cards (J, Q, K): 12, 6 of them red. Honours (A, K, Q, J): 16. Number cards (2 to 10): 36.
For 'or', subtract the overlap: a diamond or a 7 is $13 + 4 - 1 = 16$ .
The bottom of a one-card answer divides 52.

For example: a diamond that is not a face card: 10 cards, so $\frac{10}{52} = \frac{5}{26}$ .

Two Cards

Drawn together: $^{52}C_{2} = 1326$ pairs. Both from a group of g cards: $\frac{g(g - 1)}{52 \times 51}$ .
One after another without replacement: $\frac{a}{52} \times \frac{b}{51}$ , where b counts what is left after the first draw.
With replacement: $\frac{a}{52} \times \frac{b}{52}$ .
'One of each', drawn together, covers both orders.

For example: both kings is $\frac{1}{221}$ without replacement and $\frac{1}{169}$ with replacement. One red and one black drawn together: $\frac{26 \times 26}{1326} = \frac{26}{51}$ .

Balls, Bags and Tickets

$^{n}C_{2} = \frac{n(n - 1)}{2}$ and $^{n}C_{3} = \frac{n(n - 1)(n - 2)}{6}$ .
Both red: $\frac{r(r - 1)}{n(n - 1)}$ . One red and one blue: $r \times b$ pairs over $^{n}C_{2}$ .
Same colour: add the 'both red', 'both blue' and other counts.
At least one red: $1 - P(\text{no red})$ .
Missing count: turn the probability into parts of the total.
Tickets numbered 1 to n work like numbers from 1 to n.

For example: 3 red and 6 blue, two drawn together: both blue is $\frac{15}{36} = \frac{5}{12}$ .

Or, And, Independent Events

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ ; in set notation $\cup$ is “or”, $\cap$ is “and”, $\bar{A}$ is “not A”.
Mutually exclusive: $P(A \text{ or } B) = P(A) + P(B)$ .
Independent: $P(A \text{ and } B) = P(A) \times P(B)$ .
Two steps (a bag, then a ball): multiply along each path, then add.
Three groups, everyone in at least one: exactly one + exactly two + all three = total; exactly one + 2 times exactly two + 3 times all three = sum of the group sizes.

Two or Three People

With chances p and q: both $pq$ ; neither $(1 - p)(1 - q)$ ; at least one $1 - (1 - p)(1 - q)$ ; exactly one $p(1 - q) + q(1 - p)$ .
'The problem is solved' and 'the target is hit' mean at least one succeeds.
Two people contradict each other when exactly one tells the truth.
With three people, 'exactly two' has three cases, each with one failure.

For example: chances $\frac{1}{2}$ and $\frac{1}{6}$ : neither is $\frac{1}{2} \times \frac{5}{6} = \frac{5}{12}$ , so the problem is solved with probability $\frac{7}{12}$ .

Before the Exam

Mistakes to avoid:

the 11 sums of two dice taken as equally likely;
(2, 3) and (3, 2) counted as one outcome;
'at least one' worked as 'exactly one', or as P(none);
a red king counted twice in 'red or king';
a draw without replacement worked as with replacement;
odds given as the probability;
a leap year taken as 365 days, or 1 as a prime.

Then take the Chapter Test: 25 questions in 25 minutes, $+3$ for a right answer and $-1$ for a wrong one.

Probability