Elementary Probability

Lessons

Senior 6 · Basic Mathematics

Elementary Probability

5 lessons

Lesson 4 of 5

Dependent events and conditional probability

25 min

When the first result changes the second

A bag holds 5 red and 3 blue counters. In the last lesson the first counter was put back, so the second draw was independent of the first. Now suppose the first counter is not replaced. If it was red, the bag now holds 4 red and 3 blue, so the chance that the second is red falls from $\frac{5}{8}$ to $\frac{4}{7}$ . The second draw depends on the first. Probability "given that something has happened" is called conditional probability.

Definition

Events are dependent if the occurrence of one changes the probability of the other.
The conditional probability of $B$ given $A$ , written $P(B \mid A)$ , is the probability that $B$ happens when we know that $A$ has happened. For $P (A) > 0$ :

P(B \mid A) = \frac{P(A \cap B)}{P(A)}

Rearranged, this is the general multiplication rule, true for all events: $P(A \cap B) = P(A) \times P(B \mid A)$ .
$A$ and $B$ are independent exactly when $P(B \mid A) = P(B)$ : knowing $A$ tells us nothing about $B$ .

Why the formula makes sense

Knowing that $A$ has happened shrinks the sample space from $S$ to $A$ . Among the outcomes of $A$ , the ones where $B$ also happens are those in $A \cap B$ . For equally likely outcomes,

P(B \mid A) = \frac{n(A \cap B)}{n(A)} = \dfrac{\frac{n(A \cap B)}{n(S)}}{\frac{n(A)}{n(S)}} = \frac{P(A \cap B)}{P(A)}

(The middle step divides the top and the bottom by $n (S)$ .) On a tree diagram, the second-stage branches are always conditional probabilities: they are the probabilities given the branch you are on.

From a two-way table

A school records the results of 200 S6 students in a mock examination (illustrative figures):

	Passed	Did not pass	Total
Boarders	90	30	120
Day students	50	30	80
Total	140	60	200

To find a conditional probability from a table, restrict yourself to the row or column of the condition.

Worked example

Worked example 1: without replacement. Two counters are drawn one after the other, without replacement, from a bag of 5 red and 3 blue. Find (a) $P(\text{both red})$ ; (b) $P(\text{same colour})$ ; (c) $P(\text{second red} \mid \text{first blue})$ .

(a) $P(RR) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$

(b) $P(BB) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$ , so $P(\text{same}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}$

Compare (b) with the answer $\frac{17}{32}$ with replacement in the last lesson: not replacing makes "same colour" less likely.

Worked example

Worked example 2: from the table. One of the 200 students is chosen at random. Let $P$ = "passed" and $B$ = "boarder". Find $P (P)$ , $P(P \mid B)$ and $P(B \mid P)$ . Are "passed" and "boarder" independent?

$P(P) = \frac{140}{200} = 0.7$ .

Given a boarder, use the boarders' row only: $P(P \mid B) = \frac{90}{120} = 0.75$ .

Given a pass, use the "Passed" column only: $P(B \mid P) = \frac{90}{140} = \frac{9}{14} \approx 0.643$ .

Since $P(P \mid B) = 0.75 \ne 0.7 = P(P)$ , the events are dependent. Notice also that $P(P \mid B) \ne P(B \mid P)$ : the order of the condition matters.

A tree for two machines

Machine X makes 60% of a factory's items and 3% of its items are defective; machine Y makes 40% and 5% of its items are defective (illustrative figures). The second-stage branches are conditional probabilities: 0.03 is $P(D \mid X)$ and 0.05 is $P(D \mid Y)$ .

Worked example

Worked example 3: turning a tree round. For the factory above, an item is chosen at random. Use the tree diagram.

(a) Find the probability that it is defective. The two paths ending in D are mutually exclusive, so add them:

P(D) = 0.6 \times 0.03 + 0.4 \times 0.05 = 0.018 + 0.020 = 0.038

(b) The item is defective. Find the probability that it was made by machine X.

P(X \mid D) = \frac{P(X \cap D)}{P(D)} = \frac{0.018}{0.038} = \frac{9}{19} \approx 0.474

Although X makes more items, fewer than half of the defective ones come from X, because X is more reliable.

Warning

$P(B \mid A)$ is not $P(A \mid B)$ . The probability that a student who passed is a boarder (0.643) is not the probability that a boarder passed (0.75). Always ask: "Which group am I restricted to?" That group goes in the denominator.

Tip

Without replacement, reduce both the number of that kind and the total for the second draw: $\frac{5}{8}$ then $\frac{4}{7}$ , never $\frac{4}{8}$ .

Khan Academy: "Conditional probability tree diagram example". Watch how the second-stage branches are conditional probabilities, as in Worked example 3.

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Practice

Two cards are drawn from a pack of 52 without replacement. Find the probability that both are aces.
$P (A) = 0.5$ and $P(A \cap B) = 0.2$ . Find $P(B \mid A)$ .
From the mock-examination table, find (a) $P(\text{did not pass} \mid \text{day student})$ ; (b) $P(\text{day student} \mid \text{did not pass})$ .
A box has 6 good and 4 faulty bulbs. Two are taken without replacement. Find the probability that (a) both are faulty; (b) exactly one is faulty.
$P (A) = 0.4$ , $P (B) = 0.5$ and $P(A \mid B) = 0.6$ . Find $P(A \cap B)$ and $P(B \mid A)$ . Are $A$ and $B$ independent?
In the factory example, find $P(Y \mid D)$ and check that $P(X \mid D) + P(Y \mid D) = 1$ .
Explain in words why drawing without replacement makes the draws dependent.

Show answersHide answers

$\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$ .
$\frac{0.2}{0.5} = 0.4$ .
(a) $\frac{30}{80} = \frac{3}{8}$ (b) $\frac{30}{60} = \frac{1}{2}$ .
(a) $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$ (b) $\frac{6}{10} \times \frac{4}{9} + \frac{4}{10} \times \frac{6}{9} = \frac{48}{90} = \frac{8}{15}$ .
$P(A \cap B) = 0.5 \times 0.6 = 0.3$ ; $P(B \mid A) = \frac{0.3}{0.4} = 0.75$ . Not independent, since $P(A \mid B) = 0.6 \ne 0.4 = P(A)$ .
$P(Y \mid D) = \frac{0.020}{0.038} = \frac{10}{19} \approx 0.526$ ; $\frac{9}{19} + \frac{10}{19} = 1$ .
The first item removed changes what is left in the bag, so the probabilities for the second draw depend on what the first was.

Important

$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$ and $P(A \cap B) = P(A)\,P(B \mid A)$ . Events are independent exactly when $P(B \mid A) = P(B)$ .

Sources and credits

Text, table, tree diagram, worked examples and exercises: Written by Ganzaa. The examination and factory figures are illustrative.

Video: Conditional probability tree diagram example, Khan Academy, embedded from YouTube.

1When the first result changes the second

2Why the formula makes sense

3From a two-way table

4A tree for two machines

When the first result changes the second

Why the formula makes sense

From a two-way table

A tree for two machines