Probability — SPI Mathematics Formulas and Traps

Coins and a Single Die

Experiment, outcome, sample space and event

Number of outcomes

n(S) = \text{number of outcomes in the sample space}

Probability as favourable outcomes over all outcomes

Classical probability

P(E) = \frac{n(E)}{n(S)}

$n(E)$ number of outcomes favourable to the event
$n(S)$ number of outcomes in the sample space

Listing the outcomes for two or three coins

Outcomes for k coins or dice

n(S) = 2^k \text{ for } k \text{ coins} \qquad n(S) = 6^k \text{ for } k \text{ dice}

Which faces of a die are prime, composite or square

Property	Faces	Probability
Even	2, 4, 6	1/2
Odd	1, 3, 5	1/2
Prime	2, 3, 5	1/2
Odd prime	3, 5	1/3
Composite	4, 6	1/3
Perfect square	1, 4	1/3
Multiple of 3	3, 6	1/3
Neither prime nor composite	1	1/6 1 has only one factor, so it is in neither list.

Every probability is the number of faces over 6, reduced.

Common traps

Forgetting to simplify

\frac{3}{6}

is usually printed as

\frac{1}{2}

. If your fraction is not among the options, reduce it before deciding you are wrong.

'Not even' is not a new count

'Not an even number' on a die means 1, 3 or 5: the odd numbers. Rephrase a 'not' into the outcomes it leaves before counting.

Counting 1 as a prime

1 is not prime, so the primes on a die are 2, 3 and 5, giving

\frac{1}{2}

, not

\frac{2}{3}

.

Counting 2 as an odd prime

2 is prime but even. 'An odd prime' on a die is 3 or 5 only.

Treating HT and TH as one outcome

With two coins there are four outcomes, not three. Exactly one head is HT or TH, so its probability is

\frac{2}{4} = \frac{1}{2}

, not

\frac{1}{3}

.

'At most one' includes none

At most one tail with three coins means no tail or one tail: HHH, HHT, HTH, THH. Leaving out HHH gives

\frac{3}{8}

, a printed wrong option.

Probability Rules and the Complement

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Every probability lies from 0 to 1

Range of a probability

0 \le P(E) \le 1

The probability of an event not happening

Complement rule

P(\text{not } E) = 1 - P(E)

Common traps

Reading a percentage as a whole number

5% is 0.05, a valid probability. As a percentage a probability can be any value from 0 to 100, including decimals such as 12.5%.

Allowing negative probabilities

-1 \le P(E) \le 1

is a printed wrong option. A probability is never negative, so the range starts at 0.

Subtracting from 100 in the wrong column

100% minus 57% is 43%, not 53%. Write the subtraction down: a slip in the tens digit is exactly what the wrong options print.

Answering with the event instead of its complement

When P(E) = 0.57 is given, 0.57 is often among the options. The question asks for 'not E', so the answer is the other part of 1.

Two Dice and Calendar Pairs

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The 36 outcomes of two dice

Ways to make a sum s with two dice

\text{ways}(s) = 6 - |s - 7|, \quad 2 \le s \le 12

Both dice at once: multiply, or use the complement

Two independent dice

P(A \text{ and } B) = P(A) \times P(B) \qquad P(\text{at least one}) = 1 - P(\text{none})

Pairs of days: 53 Sundays and two visitors

Two people choosing from d days

P(\text{same day}) = \frac{d}{d^2} = \frac{1}{d} \qquad P(\text{different days}) = 1 - \frac{1}{d}

Common traps

Counting (2, 5) and (5, 2) once

The dice are different, so every unequal pair appears twice in the 36. Only a doublet such as (3, 3) appears once.

Counting a sum twice in 'divisible by 4 or 6'

12 is divisible by both 4 and 6. List the sums that qualify (4, 6, 8, 12) and count each one once:

3 + 5 + 5 + 1 = 14

ways.

'Between' leaves out the end values

A product between 8 and 13 means 9, 10 or 12. Counting 8 as well adds two ways and gives a printed wrong option.

Adding instead of using the complement

At least one six is not

\frac{1}{6} + \frac{1}{6} = \frac{12}{36}

: that counts (6, 6) twice. The right count is

1 - \frac{25}{36} = \frac{11}{36}

.

An even product needs only one even number

The product of two dice is even unless both are odd. Counting only 'both even' gives

\frac{1}{4}

instead of

\frac{3}{4}

.

Answering 1/7 for a leap year

A leap year has two extra days, and a given weekday falls in two of the seven possible pairs, so the answer is

\frac{2}{7}

.

\frac{1}{7}

is the ordinary-year answer.

Cards, Bags and Numbered Cards

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Bags of balls, numbered cards and letters of a word

Drawing one item

P = \frac{\text{items that fit}}{\text{all items}} \qquad \text{numbers } a \text{ to } b: \; b - a + 1 \text{ cards}

Drawing two cards or balls together

Two drawn together

P(\text{both of a kind}) = \frac{a}{n} \times \frac{a-1}{n-1}

Odds in favour and odds against

Odds and probability

\text{odds in favour} = f : u, \quad P(E) = \frac{f}{f + u}

$f$ number of favourable outcomes
$u$ number of unfavourable outcomes

The make-up of a 52-card deck

Cards	How many	Probability of one
Any one rank (all the 6s, all the queens)	4	1/13
One suit (all spades)	13	1/4
Red cards or black cards	26	1/2
Face cards (jack, queen, king)	12	3/13
Red queens, or black kings	2	1/26
One named card (the 7 of spades)	1	1/52
Aces	4	1/13 An ace is not a face card.

Probability is the count over 52, reduced.

Common traps

All the kings, or only the black ones

There are 4 kings but 2 black kings. A black king is

\frac{2}{52} = \frac{1}{26}

;

\frac{1}{13}

is any king.

Counting the ace as a face card

Face cards are the jack, queen and king only: 12 of them, so P(face card)

= \frac{3}{13}

, not

\frac{4}{13}

.

Cards from a to b are b minus a plus 1

Cards 2 to 11 are 10 cards, not 9. Subtracting the ends drops one card and changes every probability on the page.

Counting a repeated letter once

A letter drawn from a word is one of all its letters, repeats included. ASSASSINATION has three A's, two I's and one O: six vowels out of 13.

Keeping 52 in the second fraction

After one card is drawn, 51 remain.

\frac{4}{52} \times \frac{4}{52}

is the answer for drawing with replacement, which 'drawn together' is not.

Odds are not a probability

Odds of 3 : 10 mean a probability of

\frac{3}{13}

, not

\frac{3}{10}

. Add the two parts to get the total.

Counting the ace of spades twice

'A spade or an ace' is 16 cards, not 17: the ace of spades is already among the 13 spades.

Experimental Probability and Expected Counts

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Experimental probability from a record of trials

Experimental probability

P(E) = \frac{\text{times } E \text{ happened}}{\text{total trials}}

Expected number from a probability

Expected count

\text{expected number} = P(E) \times \text{total}

Common traps

Answering 1/2 for a recorded coin

When the question gives the number of heads in a record of tosses, the answer comes from that record. A fair coin's

\frac{1}{2}

is the classical value, not the experimental one.

Giving the count of the event, not the one asked

If P(bad egg) is 0.045 in 400 eggs, the bad eggs number 18. A question asking for the good eggs wants

400 - 18 = 382

, and 18 is printed as a wrong option.

Dropping a decimal place

0.055 \times 600 = 33

, not 3.3 or 330. Check the size: 0.055 is about one in twenty, and one twentieth of 600 is 30.

Probability formulas

Coins and a Single Die

Experiment, outcome, sample space and event

Probability as favourable outcomes over all outcomes

Listing the outcomes for two or three coins

Which faces of a die are prime, composite or square

Common traps

Probability Rules and the Complement

Every probability lies from 0 to 1

The probability of an event not happening

Common traps

Two Dice and Calendar Pairs

The 36 outcomes of two dice

Both dice at once: multiply, or use the complement

Pairs of days: 53 Sundays and two visitors

Common traps

Cards, Bags and Numbered Cards

Bags of balls, numbered cards and letters of a word

Drawing two cards or balls together

Odds in favour and odds against

The make-up of a 52-card deck

Common traps

Experimental Probability and Expected Counts

Experimental probability from a record of trials

Expected number from a probability

Common traps

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