SPI Mathematics · Probability

Cards, Bags and Numbered Cards

Count the cards, balls or letters that fit and divide by how many there are; for a 52-card deck, know its make-up by heart.

All Probability notes

Why this matters

Eighteen questions in the bank, the largest page, including the 2022 paper's question on drawing two queens. Half are a single card from a deck, the rest are bags of coloured balls, numbered cards and letters of a word, plus one on odds.

Concept 1 of 4: The make-up of a 52-card deck

A deck is a 4 by 13 grid: four suits, and thirteen ranks in each. Every card question is a count of cells in that grid.

Definition

Four suits of 13: spades and clubs are black, hearts and diamonds are red.
Each suit runs ace, 2 to 10, jack, queen, king, so there are 4 of every rank.
Face cards are the jack, queen and king: $3 \times 4 = 12$ .

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.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Probability · Cards, Bags and Numbered Cards

One card is drawn from a well shuffled deck of 52 cards. Then the probability of getting a queen of red suit is

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}

Drill 7 more on the make-up of a 52-card deck

Concept 2 of 4: Bags of balls, numbered cards and letters of a word

Whatever is drawn, the method is the deck's: count everything in the bag, count what fits, divide. When a count is unknown, the probabilities give an equation for it.

Definition

Total first: 6 red, 8 white, 4 green and 7 black balls make 25.
'Neither green nor white' is red or black. Either count those, or take green-or-white from 1.
Cards numbered 6 to 15 are $15 - 6 + 1 = 10$ cards: count both ends.
A letter chosen from a word: count repeated letters every time they appear (ASSASSINATION has 13 letters).
Unknown count: if P(blue) is twice P(red), the bag holds twice as many blue balls as red, because both share the same total.

Drawing one item

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

Worked example

A bag has 4 red, 5 blue and 3 green balls. One is drawn at random. Find the probability that it is not blue.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Probability · Cards, Bags and Numbered Cards

A bag contains 6 red, 8 white, 4 green and 7 black balls. One ball is drawn at random. The probability that the ball is drawn is neither green nor white is

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.

Drill 7 more on bags of balls, numbered cards and letters of a word

Concept 3 of 4: Drawing two cards or balls together

Drawing two together is drawing one and then another without putting the first back. The second draw has one fewer item in total and, if the first matched, one fewer that match.

Definition

Both items of a kind: $\frac{a}{n} \times \frac{a - 1}{n - 1}$ , where $a$ items of $n$ are of that kind.
The second fraction shrinks top and bottom: the first item is not put back.
The same answer comes from choosing pairs: $\frac{a(a-1)/2}{n(n-1)/2}$ .

Two drawn together

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

Worked example

Two cards are drawn together from a deck of 52. Find the probability that both are red.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q20Easy

Example 3 · Probability · Cards, Bags and Numbered Cards

If from a pack of 52 cards, two cards are drawn together, at random, the probability of both cards being queens is ____.

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.

Concept 4 of 4: Odds in favour and odds against

Odds compare the outcomes that fit with the ones that do not, instead of with all of them. A probability of

\frac{1}{4}

is odds of 1 to 3 in favour and 3 to 1 against.

Definition

Odds in favour $=$ favourable : unfavourable. Odds against $=$ unfavourable : favourable.
From odds $a : b$ in favour, $P(E) = \frac{a}{a + b}$ .
'A spade or an ace' counts the ace of spades once: $13 + 4 - 1 = 16$ cards.

Odds and probability

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

fnumber of favourable outcomes
unumber of unfavourable outcomes

Worked example

A card is drawn from a deck of 52. Find the odds in favour of a face card.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Probability · Cards, Bags and Numbered Cards

A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. Then, the odds against his winning the bet is

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.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

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

Reference tables (1)

The make-up of a 52-card deck7 rows

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.

Watch out for (7)

All the kings, or only the black ones
→ The make-up of a 52-card deck
Counting the ace as a face card
→ The make-up of a 52-card deck
Cards from a to b are b minus a plus 1
→ Bags of balls, numbered cards and letters of a word
Counting a repeated letter once
→ Bags of balls, numbered cards and letters of a word
Keeping 52 in the second fraction
→ Drawing two cards or balls together
Odds are not a probability
→ Odds in favour and odds against
Counting the ace of spades twice
→ Odds in favour and odds against

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