Chapter 13 · Probability

Example 8: Urn: two black balls drawn without replacement

ExamplesMultiplication rule · Section 13.3

The question

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Exam answer

Let $E$ ⁠: the first ball drawn is black, and $F$ ⁠: the second ball drawn is black. We have to find $P(E\cap F)$ ⁠.

At the first draw there are 15 balls, 10 of them black.

P(E) = \frac{10}{15}

Given that the first ball is black (not replaced), 9 black and 5 white balls are left.

P(F|E) = \frac{9}{14}

By the multiplication rule of probability,

\begin{aligned} P(E\cap F) &= P(E)\,P(F|E) \\ &= \frac{10}{15}\times\frac{9}{14} \end{aligned}

$\therefore$ P(both balls black) $= \dfrac{3}{7}$ ⁠.

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

The one idea this problem is built on

For "this, then that", multiply the first probability by the second probability given the first. Without replacement, the first black ball changes what is left in the urn, so the second draw is a conditional probability:

P(E \cap F) = P(E)\,P(F|E).

Given, to find, and the plan

Given: An urn with 10 black and 5 white balls (15 in all). Two balls are drawn one after the other, without replacement, each draw at random from the balls then in the urn.
To find: The probability that both balls are black.
Plan: Name $E$ (first black) and $F$ (second black); find $P(E)$ ⁠; find $P(F|E)$ from the urn after one black ball is removed; multiply.

Basics you need

1Equally likely draws

A ball drawn at random from $N$ balls of which $k$ are black is black with probability $\frac{k}{N}$ ⁠. Basics: the classical definition

2Conditional probability

$P(F|E)$ is the probability of $F$ when we know that $E$ has happened (Section 13.2). Lesson: conditional probability

3The multiplication rule

$P(E \cap F) = P(E)\,P(F|E)$ ⁠, provided $P(E) \ne 0$ (Section 13.3). Lesson: the multiplication rule

4Tree diagrams

Multiply along a path. Basics: tree diagrams

Solution

Name the events

Let $E$ ⁠: the first ball drawn is black, and $F$ ⁠: the second ball drawn is black. "Both black" is $E \cap F$ (also written $EF$ ⁠), so we have to find $P(E \cap F)$ ⁠.

Splitting "both black" into one event per draw lets us handle the draws one at a time.

Basics: events as sets
The first draw
$P(E) = \frac{10}{15}.$
At the first draw there are 15 balls, 10 of them black, each equally likely to be drawn.

Basics: the classical definition
The second draw, given the first was black
$P(F|E) = \frac{9}{14}.$
If $E$ has happened, one black ball is out of the urn and is not put back. That leaves 9 black and 5 white balls, 14 in all. The probability that the second ball is black, given that the first was black, is therefore $\frac{9}{14}$ ⁠.

Lesson: conditional probability
Multiply
$\begin{aligned} P(E \cap F) &= P(E)\,P(F|E) \\ &= \frac{10}{15} \times \frac{9}{14} = \frac{90}{210} = \frac{3}{7}. \end{aligned}$

The tree for the two draws. Each second-stage probability is conditional on the first draw. The highlighted path, black then black, has probability $\frac{10}{15} \times \frac{9}{14} = \frac{3}{7}$ ⁠; the four paths add up to 1.

The multiplication rule says that the chance of " $E$ and then $F$ " is the chance of $E$ times the chance of $F$ once $E$ is known; it comes from the definition $P(F|E) = \frac{P(E \cap F)}{P(E)}$ by multiplying both sides by $P(E)$ ⁠. Here $P(E) = \frac{10}{15} \ne 0$ ⁠. To simplify, $\frac{90}{210}$ ⁠: divide top and bottom by 30.

Multiplication rule, Section 13.3. Lesson: the multiplication rule

Result

P(both balls black) $= P(E \cap F) = \dfrac{3}{7}$ ⁠.

(Textbook answer: $\frac{3}{7}$ ⁠.)

Check it with numbers

By counting ordered pairs. There are $15 \times 14 = 210$ ordered ways to draw two balls, all equally likely, and $10 \times 9 = 90$ of them give black then black. So the probability is $\frac{90}{210} = \frac{3}{7}$ ⁠, the same.

The tree adds up. $\frac{3}{7} + \frac{5}{21} + \frac{5}{21} + \frac{2}{21} = \frac{9 + 5 + 5 + 2}{21} = 1$ ⁠.

Decimals. $\frac{3}{7} \approx 0.4286$ ⁠. With replacement the answer would be $\frac{10}{15} \times \frac{10}{15} = \frac{4}{9} \approx 0.4444$ ⁠, a little more: removing a black ball makes a second black slightly less likely.

Common mistakes

Using $\frac{10}{15}$ twice. That is the "with replacement" answer $\frac{4}{9}$ ⁠; here the first ball is not put back.
Changing only the black count. After one black ball is removed, the total also drops to 14: $\frac{9}{14}$ ⁠, not $\frac{9}{15}$ ⁠.
Adding instead of multiplying. "Both" means one thing and the other, so the probabilities along the path are multiplied.

What to take away

$P(E \cap F) = P(E)\,P(F|E)$ ⁠: first stage, times second stage given the first.
"Without replacement" changes both the count of the colour drawn and the total. The same reasoning with cards is Exercise 13.2, Question 2.
A tree diagram shows every case; its path probabilities add up to 1.

Every fact used, and where it comes from

Fact	Used in	Source
Probability of a black ball is $\frac{k}{N}$ for equally likely draws	Steps 2, 3	Class XI, Probability (Basics: the classical definition)
Conditional probability $P(F\|E)$	Step 3	Definition 1, Section 13.2 (Lesson: conditional probability)
$P(E \cap F) = P(E)\,P(F\|E)$ ⁠, $P(E) \ne 0$	Step 4	Multiplication rule, Section 13.3 (Lesson: the multiplication rule)
Tree diagram: multiply along a path, paths add to 1	Step 4, Check	Basics: tree diagrams
Ordered draws: $15 \times 14$ and $10 \times 9$	Check	Counting principle, Class XI (Basics: multiply the choices)

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Example 8: Urn: two black balls drawn without replacement

Basics you need

Solution

Name the events

The first draw

The second draw, given the first was black

Multiply

Every fact used, and where it comes from