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The Lumenary · Grades 9–12

Conditional probability and independence

Name: ______________________

Date: ______________________

You will learn to

Compute conditional probabilities from two-way tables and tree diagrams, and use the multiplication rule
Decide whether events are independent, and distinguish independence from mutual exclusivity
Prove the rules of conditional probability, the law of total probability and Bayes' theorem from the definitions
Apply Bayes' theorem to diagnostic tests with base rates, using natural frequencies
Recognise the confusion of $P(A \mid B)$ with $P(B \mid A)$ , including the prosecutor's fallacy

1.
Two fair dice are rolled and you are told that the total is $9$ . What is the probability that the first die shows a $5$ ?
- A. $\tfrac14$
- B. $\tfrac16$
- C. $\tfrac{1}{36}$
- D. $\tfrac12$
2.
In the lesson's survey of $400$ students in Hanoi, $200$ cycle to school, $80$ were late at least once, and $30$ cyclists were late. A student is chosen at random from those who were late. What is the probability that they cycle?
- A. $\tfrac{30}{80} = 0.375$
- B. $\tfrac{30}{200} = 0.15$
- C. $\tfrac{30}{400} = 0.075$
- D. $\tfrac{200}{400} = 0.5$
3.
A condition affects $2\%$ of a population of $10\,000$ people. A test detects it in $90\%$ of people who have it, and gives a false positive for $10\%$ of people who do not. If everyone is tested, how many positive results are expected?
4.
With the numbers of the previous exercise (base rate $2\%$ , $90\%$ detection, $10\%$ false positives), a randomly chosen person tests positive. About how likely is it that they have the condition?
- A.About $16\%$ ( $180$ out of $1160$ )
- B. $90\%$
- C. $2\%$
- D. $80\%$ , because the test is right $90\%$ of the time and wrong $10\%$
5.
Two fair dice are rolled. Which pairs of events are independent? (Choose all that apply.)
- A.“The first die shows $3$ ” and “the second die is even”.
- B.“The total is $7$ ” and “the first die shows $3$ ”.
- C.“The total is $8$ ” and “the first die shows $3$ ”.
- D.“The first die is even” and “the total is even”.
- E.“Both dice show $6$ ” and “the total is $12$ ”.
6.
Events $A$ and $B$ are mutually exclusive, with $P(A) = 0.3$ and $P(B) = 0.4$ . Which statement is true?
- A. $A$ and $B$ are not independent, because $P(A \cap B) = 0 \ne 0.12 = P(A)P(B)$ .
- B. $A$ and $B$ are independent, because they have nothing to do with each other.
- C. $P(A \cup B) = 0.3 + 0.4 - 0.12 = 0.58$ .
- D. $P(A \mid B) = 0.3$ .
7.
Two cards are drawn from a shuffled pack of $52$ without replacement. The probability that both are aces is $\dfrac{1}{N}$ . What is $N$ ?
8.
Here is the proof that if $A$ and $B$ are independent then so are $A$ and $B^c$ :

$P(A \cap B^c) \overset{(1)}{=} P(A) - P(A \cap B) \overset{(2)}{=} P(A) - P(A)P(B) \overset{(3)}{=} P(A)\big(1 - P(B)\big) \overset{(4)}{=} P(A)P(B^c)$ .

Which step uses the hypothesis that $A$ and $B$ are independent?
- A.Step (1)
- B.Step (2)
- C.Step (3)
- D.Step (4)
9.
In the lesson's Lean model, μ.cond A B is defined as μ.P (A ∩ B) / μ.P B, and Lean's convention makes x / 0 = 0. The lesson proves

theorem bayes (μ : FinProb Ω) (A B : Finset Ω) (hA : μ.P A ≠ 0) : μ.cond A B = μ.cond B A * μ.P A / μ.P B.

What is the role of the hypothesis hA?
- A.It is needed so that μ.cond B A * μ.P A equals μ.P (B ∩ A), the step where the multiplication rule is used.
- B.It prevents division by zero in μ.P B; without it Lean would crash.
- C.None: it is decoration and could be deleted without changing anything.
- D.It says that A and B are independent.
10.
Write posterior(prior, sensitivity, false_positive_rate) returning $P(D \mid \text{positive})$ by Bayes' theorem, and after_positives(prior, sensitivity, false_positive_rate, k) returning the probability of $D$ after $k$ positive results from independent tests (use each posterior as the next prior). The tests pass Fractions, so keep the arithmetic exact.
Computational — Practice

Answer key — Conditional probability and independence

1.
Given the total is $9$ , the outcomes are $(3,6), (4,5), (5,4), (6,3)$ , equally likely. One has first die $5$ , so the conditional probability is $\tfrac14$ . By the formula: $P(A \mid B) = \dfrac{1/36}{4/36} = \tfrac14$ .
2.
$P(\text{cycle} \mid \text{late}) = \dfrac{P(\text{cycle} \cap \text{late})}{P(\text{late})} = \dfrac{30/400}{80/400} = \dfrac{30}{80} = 0.375$ . Compare $P(\text{late} \mid \text{cycle}) = 0.15$ : swapping the events changes the question.
3.
$2\%$ of $10\,000$ is $200$ people with the condition, of whom $90\%$ , that is $180$ , test positive. The other $9\,800$ do not have it, and $10\%$ of them, $980$ , test positive. So $180 + 980 = 1160$ positive results are expected — this is the law of total probability, $P(T) = P(T \mid D)P(D) + P(T \mid D^c)P(D^c) = 0.9 \cdot 0.02 + 0.1 \cdot 0.98 = 0.116$ .
4.
Of the $1160$ positives, $180$ have the condition, so $P(D \mid T) = \tfrac{180}{1160} = \tfrac{9}{58} \approx 0.155$ . Bayes' theorem gives the same: $\dfrac{0.9 \times 0.02}{0.9 \times 0.02 + 0.1 \times 0.98} = \dfrac{0.018}{0.116} \approx 0.155$ .
5.
(a) $\tfrac{3}{36} = \tfrac16 \cdot \tfrac12$ : independent. (b) $\tfrac{1}{36} = \tfrac16 \cdot \tfrac16$ : independent. (c) $\tfrac{1}{36} \ne \tfrac{5}{216}$ : dependent. (d) $\tfrac{9}{36} = \tfrac12 \cdot \tfrac12$ : independent. (e) the same event, $\tfrac{1}{36} \ne \tfrac{1}{1296}$ : dependent.
6.
$P(A \cap B) = 0$ because the events cannot both happen, while $P(A)P(B) = 0.12$ . So they are not independent; indeed $P(A \mid B) = 0$ . Also $P(A \cup B) = 0.3 + 0.4 = 0.7$ .
7.
$P = \tfrac{4}{52} \times \tfrac{3}{51} = \tfrac{12}{2652} = \tfrac{1}{221}$ , so $N = 221$ . Check by counting: $\binom{4}{2} = 6$ pairs of aces among $\binom{52}{2} = 1326$ pairs of cards, and $\tfrac{6}{1326} = \tfrac{1}{221}$ .
8.
Step (2). Step (1) is additivity for the disjoint pieces $A \cap B$ and $A \cap B^c$ ; step (3) is algebra; step (4) is the complement rule. Only step (2) uses $P(A \cap B) = P(A)P(B)$ . In Lean this is indep_compl, where the hypothesis h is used once, in the same place.
9.
The proof is rw [← mul_rule μ B A hA, inter_comm, cond]: it replaces μ.cond B A * μ.P A by μ.P (B ∩ A), and mul_rule needs μ.P A ≠ 0 for that. This mirrors the human statement, where $P(B \mid A)$ is only defined when $P(A) > 0$ .

10.

from fractions import Fraction


def posterior(prior, sensitivity, false_positive_rate):
    top = sensitivity * prior
    return top / (top + false_positive_rate * (1 - prior))


def after_positives(prior, sensitivity, false_positive_rate, k):
    p = prior
    for _ in range(k):
        p = posterior(p, sensitivity, false_positive_rate)
    return p

One positive result raises $1\%$ to $\tfrac{19}{118} \approx 16\%$ ; a second independent positive raises it to about $78.5\%$ . Using the posterior as the next prior is valid only if the tests' errors are independent given the person's true state.