Chapter 13 · Probability

Exercise 13.3 · Q1: Urn topped up with the colour drawn: second ball red, $P = \frac12$

Exercise 13.3Theorem of total probability · Section 13.5

The question

An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Exam answer

Let $E_1$ ⁠: the first ball drawn is red, $E_2$ ⁠: the first ball drawn is black, and $A$ ⁠: the second ball drawn is red.

$E_1$ and $E_2$ form a partition of the sample space (the first ball is either red or black).

$P(E_1) = \frac{5}{10} = \frac12$ and $P(E_2) = \frac{5}{10} = \frac12$ ⁠.

If $E_1$ occurs, the urn has 7 red and 5 black balls, so $P(A|E_1) = \frac{7}{12}$ ⁠.

If $E_2$ occurs, the urn has 5 red and 7 black balls, so $P(A|E_2) = \frac{5}{12}$ ⁠.

By the theorem of total probability,

\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac12\cdot\frac{7}{12} + \frac12\cdot\frac{5}{12} \\ &= \frac{7}{24} + \frac{5}{24} = \frac{12}{24} \end{aligned}

$\therefore$ The probability that the second ball is red is $\dfrac12$ ⁠.

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

The one idea this problem is built on

The second draw depends on what the first draw was, so split by the first ball: it is either red or black, and exactly one of these happens. For each case the urn is known exactly, so the chance of a red second ball is easy. The theorem of total probability then adds the two cases, each weighted by how likely it is: multiply along each branch of the tree and add the branches that end in "red".

Given, to find, and the plan

Given: An urn with 5 red and 5 black balls. The first ball is put back together with 2 more balls of its colour, so the urn then has 12 balls.
To find: The probability that the second ball drawn is red.
Plan: (1) Name the events and check that the two "first ball" events form a partition. (2) Find $P(E_1)$ and $P(E_2)$ ⁠. (3) Count the urn after each case to get $P(A|E_1)$ and $P(A|E_2)$ ⁠. (4) Apply the theorem of total probability.

Basics you need

1Equally likely outcomes

When every ball is equally likely to be drawn, $P(E) = \frac{n(E)}{n(S)}$ ⁠: the number of favourable balls over the total number of balls. Basics: classical probability

2Conditional probability (Section 13.2)

$P(A|E)$ is the probability of $A$ once we know that $E$ has happened. Here it is simply the chance of red in the urn as it stands after the first draw. Lesson: conditional probability

3Partition (Section 13.5.1)

Events $E_1, E_2, \ldots, E_n$ form a partition of the sample space $S$ when they are pairwise disjoint, together make up $S$ ⁠, and each has nonzero probability. Lesson: total probability and Bayes' theorem

4Theorem of total probability (Section 13.5.2)

If $E_1, E_2$ form a partition, then $P(A) = P(E_1)P(A|E_1) + P(E_2)P(A|E_2)$ ⁠. On a tree: multiply along each path, then add the paths that end in $A$ ⁠. Basics: tree diagrams

Solution

Name the events

Let
- $E_1$ ⁠: the first ball drawn is red;
- $E_2$ ⁠: the first ball drawn is black;
- $A$ ⁠: the second ball drawn is red.
$E_1$ and $E_2$ form a partition of the sample space: they cannot both happen (the first ball has one colour), one of them must happen (every ball is red or black), and each has a nonzero probability (Step 2).

The second draw is not the same experiment whatever happens first: the urn changes. Splitting by the first ball turns one hard question into two easy ones, and the partition is exactly the condition that lets the theorem of total probability add them back together.

A partition: pairwise disjoint, exhaustive, nonzero probabilities (Section 13.5.1). Lesson: total probability
The chances for the first ball

The urn starts with $5 + 5 = 10$ balls, all equally likely to be drawn, and 5 of them are red. So
$\begin{aligned} P(E_1) &= \frac{5}{10} = \frac12, \\ P(E_2) &= \frac{5}{10} = \frac12. \end{aligned}$
"Drawn at random" means every ball has the same chance, so the classical formula applies. The two probabilities add up to 1, as they must for a partition into two events.

$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: classical probability
The urn after each first draw

The first ball goes back, and 2 more balls of its colour are added, so the urn always holds $10 + 2 = 12$ balls for the second draw.
- If the first ball was red ( $E_1$ ⁠): 7 red and 5 black, so $P(A|E_1) = \frac{7}{12}$ ⁠.
- If the first ball was black ( $E_2$ ⁠): 5 red and 7 black, so $P(A|E_2) = \frac{5}{12}$ ⁠.
Knowing which case happened tells us the exact contents of the urn, and the second ball is drawn at random from those 12 balls. The returned ball counts too: forgetting it would give 11 balls.

$P(A|E)$ ⁠: the chance of $A$ in the situation that $E$ leaves behind (Section 13.2). Lesson: conditional probability
Add the two routes: the theorem of total probability

Since $E_1$ and $E_2$ form a partition, the theorem of total probability gives
$\begin{aligned} P(A) &= P(E_1)P(A|E_1) \\ &\quad + P(E_2)P(A|E_2) \\ &= \frac12\cdot\frac{7}{12} + \frac12\cdot\frac{5}{12} \\ &= \frac{7}{24} + \frac{5}{24} \\ &= \frac{12}{24} = \frac12. \end{aligned}$

The tree of the two draws. Multiply along each path; the two green paths end in a red second ball, and their products add to $\frac{7}{24} + \frac{5}{24} = \frac12$ ⁠.

Each term is one route to a red second ball: $P(E_1)P(A|E_1) = P(E_1 \cap A)$ by the multiplication rule, and similarly for $E_2$ ⁠. The two routes cannot happen together, so their probabilities add.

Multiply along a path, add the paths. Basics: tree diagrams
State the answer, and see why it is so neat

The probability that the second ball is red is $\frac12$ ⁠.

It is the same as for the first ball. Adding balls of the colour drawn helps red after a red draw ( $\frac{7}{12}$ ⁠) exactly as much as it helps black after a black draw, and the urn starts balanced, so the two effects cancel.

The second ball is also black with probability $\frac{5}{24} + \frac{7}{24} = \frac12$ ⁠, and the two answers add to 1, as they must.

$P(A') = 1 - P(A)$ ⁠. Basics: rules of probability

Result

The probability that the second ball is red is $\frac12$ ⁠. (Textbook answer: $\frac12$ ⁠.)

Check it with numbers

Give every ball a name and list every way the two draws can go: 10 choices for the first ball, then 12 for the second, all equally likely, so $10 \times 12 = 120$ outcomes. A red second ball happens in $5 \times 7 = 35$ of them after a red first ball and $5 \times 5 = 25$ after a black one: $35 + 25 = 60$ outcomes, and $\frac{60}{120} = 0.5$ ⁠. The verify script makes this list and counts it.

Common mistakes

Forgetting the returned ball. The urn has $10 + 2 = 12$ balls for the second draw, not 11.
Answering $\frac{7}{12}$ ⁠. That is $P(A|E_1)$ ⁠, the chance only when the first ball was red. The first ball may be black, so both cases must be weighted and added.
Adding $\frac{7}{12} + \frac{5}{12}$ ⁠. Each conditional probability must first be multiplied by the probability of its case.
Treating the draws as independent. The second draw depends on the first, because the urn changes.

What to take away

When the second stage depends on the first, split by the first stage (a partition), find the conditional probability in each case, and add with the theorem of total probability.
With $r$ red and $b$ black balls and $c$ balls of the drawn colour added, the second ball is red with probability $\frac{r}{r + b}$ ⁠, the same as the first. Here $\frac{5}{10} = \frac12$ ⁠.
The next questions turn the question round: given the result, which case caused it? That is Bayes' theorem, starting in Question 2.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Steps 2, 3	Classical probability, Class XI (Basics: classical probability)
Conditional probability $P(A\|E)$	Step 3	Section 13.2 (Lesson: conditional probability)
Multiplication rule $P(E \cap A) = P(E)P(A\|E)$	Step 4	Section 13.3 (Lesson: the multiplication rule)
Partition of a sample space	Step 1	Section 13.5.1 (Lesson: total probability)
Theorem of total probability	Step 4	Section 13.5.2; used in Example 15 (Examples 15 to 21)
$P(A') = 1 - P(A)$	Step 5	Class XI axioms (Basics: rules of probability)

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Exercise 13.3 · Q1: Urn topped up with the colour drawn: second ball red, P=12P = \frac12P=21​

Basics you need

Solution

Name the events

The chances for the first ball

The urn after each first draw

Add the two routes: the theorem of total probability

State the answer, and see why it is so neat

Every fact used, and where it comes from

Exercise 13.3 · Q1: Urn topped up with the colour drawn: second ball red, $P = \frac12$