Chapter 13 · Probability

Exercise 13.2 · Q2: Two cards without replacement: both black

Exercise 13.2Multiplication rule of probability · Section 13.3

The question

Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Exam answer

Let $E$ be the event that the first card is black and $F$ the event that the second card is black.

A pack has 26 black cards out of 52.

P(E) = \frac{26}{52} = \frac{1}{2}

After a black card is drawn (not replaced), 25 black cards are left out of 51.

P(F|E) = \frac{25}{51}

By the multiplication rule of probability,

\begin{aligned} P(E\cap F) &= P(E)\,P(F|E) \\ &= \frac{1}{2}\times\frac{25}{51} \end{aligned}

$\therefore$ P(both cards black) $= \dfrac{25}{102}$ ⁠.

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

The one idea this problem is built on

When the first card is not put back, the second draw happens in a smaller, changed pack. So we cannot just square $\frac12$ ⁠. The multiplication rule handles this: the probability that both happen is the probability of the first, times the probability of the second given the first, $P(E\cap F) = P(E)\,P(F|E)$ ⁠.

Given, to find, and the plan

Given: A pack of 52 cards: 26 black (spades and clubs) and 26 red. Two cards are drawn one after the other, the first not replaced.
To find: The probability that both cards are black.
Plan: (1) Name the events for the first and second draw. (2) Find $P(E)$ ⁠. (3) Find $P(F|E)$ by looking at the pack that is left. (4) Multiply. (5) Check by counting pairs of cards with combinations.

Basics you need

1Classical probability

When all outcomes are equally likely, $P(E) = \frac{n(E)}{n(S)}$ ⁠: favourable outcomes over all outcomes. Basics: P(E) = n(E)/n(S)

2Conditional probability

$P(F|E)$ is the probability of $F$ once we know $E$ has happened. For a draw, that means: look at the cards that are left after the first draw. Lesson: conditional probability

3Multiplication rule

$P(E\cap F) = P(E)\,P(F|E)$ ⁠, provided $P(E) \ne 0$ (Section 13.3). On a tree diagram: multiply along the path. Basics: tree diagrams

4The pack of cards

52 cards in four suits of 13: spades and clubs are black, hearts and diamonds are red. So there are $2\times 13 = 26$ black cards.

Solution

Name the events

Let $E$ be the event "the first card drawn is black" and $F$ the event "the second card drawn is black". "Both cards are black" is the event $E\cap F$ ⁠.

Splitting "both black" into one event per draw lets us handle the draws one at a time, which is exactly what the multiplication rule needs. The book does the same with the two black balls of Example 8.

$E\cap F$ means " $E$ and $F$ both happen". Basics: events as sets
The first draw: $P(E)$

All 52 cards are equally likely to come first, and 26 of them are black:
$P(E) = \frac{26}{52} = \frac{1}{2}.$
This is the classical definition: 26 favourable cards out of 52 equally likely cards. Half the pack is black, so the answer $\frac12$ makes sense.

$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: classical probability
The second draw, given the first was black: $P(F|E)$

If $E$ has happened, one black card has gone and is not replaced. The pack now has $51$ cards, of which $26 - 1 = 25$ are black. So
$P(F|E) = \frac{25}{51}.$
Knowing $E$ shrinks the world: the second card is drawn from the 51 cards that are left, and each of them is equally likely. Because the black card was not put back, the second draw is slightly less likely to be black than the first ( $\frac{25}{51} < \frac12$ ⁠). This change is why $E$ and $F$ are not independent here.

Conditional probability as "the probability in the reduced sample space" (Section 13.2). Lesson: conditional probability
Multiply along the path

By the multiplication rule of probability,
$\begin{aligned} P(E\cap F) &= P(E)\,P(F|E) \\ &= \frac{1}{2}\times\frac{25}{51} \\ &= \frac{25}{102}. \end{aligned}$
$25 = 5^2$ and $102 = 2\times 3\times 17$ share no factor, so the fraction is in lowest terms.

The tree of the two draws (B = black, R = red). The probabilities on the second branches depend on the first card, because it is not put back. The highlighted path is "black, then black": $\frac{26}{52}\times\frac{25}{51} = \frac{25}{102}$ ⁠.

The rule says: to get "first black and second black", take the chance of the first and scale it by the chance of the second in the situation the first has left behind. On the tree that is multiplying the two probabilities along the highlighted path.

Multiplication rule, Section 13.3: $P(E\cap F) = P(E)\,P(F|E)$ ⁠, $P(E) \ne 0$ ⁠. Lesson: the multiplication rule
Check by counting pairs of cards

Ignore the order and count unordered pairs. The number of ways to choose 2 cards from 52 is $^{52}C_2$ ⁠, and the number of ways to choose 2 black cards from 26 is $^{26}C_2$ ⁠:
$\begin{aligned} P &= \frac{^{26}C_2}{^{52}C_2} = \frac{325}{1326} \\ &= \frac{25\times 13}{102\times 13} = \frac{25}{102}. \end{aligned}$
Here $^{26}C_2 = \frac{26\times 25}{2} = 325$ and $^{52}C_2 = \frac{52\times 51}{2} = 1326$ ⁠. The same answer.

Drawing two cards one after the other without replacement gives every pair of cards the same chance, so counting pairs is a second, independent route. Getting $\frac{25}{102}$ twice is strong evidence that the multiplication was set up correctly.

Combinations from Class XI: $^{n}C_2 = \frac{n(n-1)}{2}$ ⁠. Multiply the choices. Basics: counting

Result

P(both cards black) $= \frac{26}{52}\times\frac{25}{51} = \frac{25}{102}$ ⁠.

(Textbook answer: $\frac{25}{102}$ ⁠.)

Check it with numbers

Count ordered draws: there are $52\times 51 = 2652$ ways to draw a first and a second card, and $26\times 25 = 650$ of them give two black cards. So the probability is $\frac{650}{2652} \approx 0.2451$ ⁠, and $\frac{25}{102} \approx 0.2451$ as well. With replacement it would be $\frac12\times\frac12 = 0.25$ ⁠; without replacement it is a little smaller, as expected, because one black card has gone.

Common mistakes

Squaring $\frac12$ ⁠. $\frac12\times\frac12 = \frac14$ treats the draws as independent, which is true only with replacement.
Forgetting that the pack shrinks. $\frac{25}{52}$ for the second draw keeps 52 cards in the denominator; only 51 are left.
Using 25 black cards and 52 for the first draw, or 26 and 51 for the second. Each fraction must describe the pack at the moment of that draw: 26 of 52 first, then 25 of 51.

What to take away

Without replacement: $P(E\cap F) = P(E)\,P(F|E)$ ⁠, with $P(F|E)$ read from the pack that is left.
Tree diagrams make the rule visual: multiply along a path.
Counting with combinations, $\frac{^{26}C_2}{^{52}C_2}$ ⁠, gives the same answer and is a good check.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 2, Check	Class XI, Probability (recalled in Section 13.1)
Conditional probability in the reduced sample space	Step 3	Section 13.2, Definition 1 (Lesson: conditional probability)
$P(E\cap F) = P(E)\,P(F\|E)$ ⁠, $P(E) \ne 0$	Step 4	Multiplication rule, Section 13.3 (Lesson: Examples 8 and 9)
Two balls drawn without replacement, same method	Step 1	Example 8, Section 13.3
$^{n}C_2 = \frac{n(n-1)}{2}$	Step 5	Class XI, Permutations and Combinations

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Exercise 13.2 · Q2: Two cards without replacement: both black

Basics you need

Solution

Name the events

The first draw: P(E)P(E)P(E)

The second draw, given the first was black: P(F∣E)P(F|E)P(F∣E)

Multiply along the path

Check by counting pairs of cards

Every fact used, and where it comes from

The first draw: $P(E)$

The second draw, given the first was black: $P(F|E)$