Chapter 13 · Probability

Example 9: Three cards: two kings, then an ace

ExamplesMultiplication rule · Section 13.3

The question

Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

Exam answer

Let $K$ ⁠: the card drawn is a king, and $A$ ⁠: the card drawn is an ace. We have to find $P(KKA)$ ⁠.

A pack has 4 kings in 52 cards.

P(K) = \frac{4}{52}

After one king is drawn, 3 kings are left in 51 cards.

P(K|K) = \frac{3}{51}

After two kings are drawn, 4 aces are left in 50 cards.

P(A|KK) = \frac{4}{50}

By the multiplication rule of probability for three events,

\begin{aligned} &P(KKA) \\ &= P(K)\,P(K|K)\,P(A|KK) \\ &= \frac{4}{52}\times\frac{3}{51}\times\frac{4}{50} \\ &= \frac{1}{13}\times\frac{1}{17}\times\frac{2}{25} \end{aligned}

$\therefore\ P(KKA) = \dfrac{2}{5525}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The multiplication rule extends to any number of stages. Multiply the probability of the first draw by the probability of each later draw given everything drawn before it:

\begin{aligned} &P(E \cap F \cap G) \\ &= P(E)\,P(F|E)\,P(G|E \cap F). \end{aligned}

Given, to find, and the plan

Given: A well shuffled pack of 52 cards (4 kings, 4 aces); three cards drawn one after another, without replacement.
To find: The probability that the first two cards are kings and the third is an ace.
Plan: Find the probability of each draw given the earlier ones: $P(K)$ ⁠, $P(K|K)$ ⁠, $P(A|KK)$ ⁠; multiply them by the multiplication rule for three events.

Basics you need

1A pack of cards

52 cards in 4 suits of 13; each of the 13 ranks (such as king or ace) appears 4 times. Basics: the classical definition

2The multiplication rule

$P(E \cap F) = P(E)\,P(F|E)$ ⁠, and for three events $P(E \cap F \cap G) = P(E)\,P(F|E)\,P(G|E \cap F)$ (Section 13.3). Lesson: the multiplication rule

3Simplifying a product of fractions

Cancel each fraction first: $\frac{4}{52} = \frac{1}{13}$ ⁠, and so on. Basics: fractions

Solution

Name the events and read the notation

Let $K$ ⁠: the card drawn is a king, and $A$ ⁠: the card drawn is an ace. We have to find $P(KKA)$ ⁠, the probability of king, then king, then ace.

The book writes the draws in order: $KKA$ is short for "first card king $\cap$ second card king $\cap$ third card ace". Likewise $P(K|K)$ means "the second card is a king, given that the first was a king", and $P(A|KK)$ means "the third card is an ace, given that the first two were kings".

Lesson: the multiplication rule
The first card
$P(K) = \frac{4}{52}.$
The pack is well shuffled, so each of the 52 cards is equally likely to be first; 4 of them are kings.

Basics: the classical definition
The second card, given a king first
$P(K|K) = \frac{3}{51}.$
One king has been drawn and not replaced: $52 - 1 = 51$ cards are left, and 3 of them are kings.

Lesson: conditional probability
The third card, given two kings first
$P(A|KK) = \frac{4}{50}.$
Two cards are gone, both kings, so 50 cards are left. No ace has been removed, so all 4 aces are still there.

Basics: the classical definition
Multiply
$\begin{aligned} P(KKA) &= P(K)\,P(K|K)\,P(A|KK) \\ &= \frac{4}{52} \times \frac{3}{51} \times \frac{4}{50} \\ &= \frac{1}{13} \times \frac{1}{17} \times \frac{2}{25} = \frac{2}{5525}. \end{aligned}$

Each draw's probability is worked out from the cards still in the pack. Multiplying the three gives $P(KKA) = \frac{2}{5525}$ ⁠.

This is the multiplication rule for three events with $E$ ⁠: first card king, $F$ ⁠: second card king, $G$ ⁠: third card ace. Cancelling first keeps the numbers small: $\frac{4}{52} = \frac{1}{13}$ ⁠, $\frac{3}{51} = \frac{1}{17}$ ⁠, $\frac{4}{50} = \frac{2}{25}$ ⁠, and $13 \times 17 \times 25 = 5525$ ⁠.

Multiplication rule for more than two events, Section 13.3. Lesson: the multiplication rule

Result

$P(KKA) = \dfrac{2}{5525}$ ⁠.

(Textbook answer: $\frac{2}{5525}$ ⁠.)

Check it with numbers

By counting ordered draws. There are $52 \times 51 \times 50 = 132600$ ordered ways to draw three cards, all equally likely. King, king, ace can happen in $4 \times 3 \times 4 = 48$ ways. So the probability is $\frac{48}{132600} = \frac{2}{5525}$ ⁠, the same.

Decimals. $\frac{2}{5525} \approx 0.000362$ ⁠, about 4 chances in 10000.

Order matters. Two kings and one ace in any order (KKA, KAK or AKK) would be three times as likely, $\frac{6}{5525}$ ⁠; the question fixes the order.

Common mistakes

Keeping 52 in every denominator. The cards are not replaced, so the pack shrinks to 51 and then 50.
Reducing the aces. Only kings have been removed before the third draw, so there are still 4 aces.
Multiplying by 3 for the orders. The question says the first two are kings and the third is an ace: only one order.

What to take away

For several draws, multiply each draw's probability given all the earlier draws.
Track two things after each draw: how many cards are left, and how many of the kind you want.
Cancel each fraction before multiplying.

Every fact used, and where it comes from

Fact	Used in	Source
A pack has 52 cards with 4 kings and 4 aces	Steps 2 to 4	Class XI, Probability (Basics: the classical definition)
$P(E \cap F \cap G) = P(E)\,P(F\|E)\,P(G\|E \cap F)$	Step 5	Multiplication rule for more than two events, Section 13.3 (Lesson: the multiplication rule)
Ordered draws: $52 \times 51 \times 50$ and $4 \times 3 \times 4$	Check	Counting principle, Class XI (Basics: multiply the choices)

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Example 9: Three cards: two kings, then an ace

Basics you need

Solution

Name the events and read the notation

The first card

The second card, given a king first

The third card, given two kings first

Multiply

Every fact used, and where it comes from