AP Statistics 2.3-2.4: Conditional Probability and Independent Events

What formula defines the conditional probability of event $A$ given event $B$?

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$, provided $P(B)>0$. The condition restricts the relevant sample space to outcomes in $B$.

Joint probability

The probability that two events occur together, written $P(A\cap B)$ or $P(A\text{ AND }B)$.

How can the probability of two events occurring together be calculated using a conditional probability?

$P(A\cap B)=P(B)P(A\mid B)=P(A)P(B\mid A)$. The order of the events does not change the intersection.

Why is $P(A\mid B)$ generally different from $P(B\mid A)$?

They condition on different events and therefore use different restricted sample spaces: $P(A\mid B)=P(A\cap B)/P(B)$, whereas $P(B\mid A)=P(A\cap B)/P(A)$.

Independent events

Events $A$ and $B$ are independent if knowing that one occurred does not change the probability of the other. For example, $P(A\mid B)=P(A)$ when $P(B)>0$.

How can independence be tested using joint probabilities?

Events $A$ and $B$ are independent if $P(A\cap B)=P(A)P(B)$. If the two sides are unequal, the events are dependent.

How can independence be tested using conditional probability?

Compare $P(A\mid B)$ with $P(A)$, or $P(B\mid A)$ with $P(B)$. Equality indicates independence, assuming the conditioning event has nonzero probability.

Mutually exclusive events

Events that cannot occur simultaneously. For mutually exclusive events, $A\cap B=\varnothing$ and therefore $P(A\cap B)=0$.

Are mutually exclusive events with positive probabilities independent?

No. If $P(A)>0$ and $P(B)>0$, then $P(A)P(B)>0$, but mutually exclusive events have $P(A\cap B)=0$, so the independence condition fails.

What is the general addition rule for the probability that at least one of two events occurs?

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$. The intersection is subtracted because it is counted in both $P(A)$ and $P(B)$.

When can the addition rule be simplified to $P(A\cup B)=P(A)+P(B)$?

When $A$ and $B$ are mutually exclusive, because then $P(A\cap B)=0$.

Why must the intersection be subtracted in the general addition rule?

Outcomes in $A\cap B$ are included once in $P(A)$ and again in $P(B)$. Subtracting $P(A\cap B)$ prevents double-counting.

Complement of an event

The complement $A'$ or $A^c$ consists of outcomes in which $A$ does not occur. Its probability is $P(A')=1-P(A)$.

How can the probability of a positive result be found if a test's negative-result probability is known?

If positive and negative are complementary outcomes, use $P(\text{positive})=1-P(\text{negative})$.

A player makes the first attempt with probability $0.65$, and the probability of making the second attempt given success on the first is $0.90$. What is the probability of two successes?

Use the multiplication rule: $P(A\cap B)=P(A)P(B\mid A)=(0.65)(0.90)=0.585$.

For the two-attempt situation with $P(A)=P(B)=0.65$ and $P(A\cap B)=0.585$, what is $P(A\cup B)$?

$P(A\cup B)=0.65+0.65-0.585=0.715$.

Why are the two attempts in the previous example not independent?

Independence would require $P(A\cap B)=(0.65)(0.65)=0.4225$, but the actual joint probability is $0.585$. Equivalently, $P(B\mid A)=0.90\ne0.65=P(B)$.

Why are success on the first attempt and success on the second attempt not mutually exclusive?

They can occur together, since $P(A\cap B)=0.585$, which is greater than zero.

Conditional probability

The probability that event $A$ occurs given that event $B$ is known or assumed to have occurred. It is written $P(A\mid B)$.

A swimmer has 150 members: 75 advanced, 47 intermediate, and the rest novice. What is the probability that a randomly selected member is novice?

There are $150-75-47=28$ novice members, so $P(\text{novice})=28/150\approx0.187$.

In a sample of 150 swimmers, 40 advanced, 30 intermediate, and 10 novice swimmers practice four times weekly. What is the probability a randomly selected swimmer practices four times weekly?

There are $40+30+10=80$ such swimmers, so the probability is $80/150\approx0.533$.

How can a contingency table be used to calculate a joint probability?

For a randomly selected individual, divide the cell count representing both characteristics by the grand total: $P(A\cap B)=\text{cell count}/\text{grand total}$.

How can a contingency table be used to calculate a conditional probability?

Restrict the denominator to the total for the condition. For example, $P(A\mid B)=\text{count in both }A\text{ and }B/\text{total count in }B$.

A table has 25 drivers who both use a phone and had a violation, 70 drivers with violations total, and 305 phone users total. Find $P(\text{phone user}\mid\text{violation})$.

The condition restricts the sample space to the 70 drivers with violations: $P(\text{phone}\mid\text{violation})=25/70\approx0.357$.

A table has 405 non-phone users with no violation and 450 non-phone users total. Find the probability of no violation given that a driver does not use a phone.

$P(\text{no violation}\mid\text{not phone})=405/450=0.900$.

How do row and column totals in a contingency table relate to the grand total?

The row totals and column totals each sum to the grand total. Dividing a marginal total by the grand total gives the corresponding unconditional or marginal probability.

Marginal probability

The unconditional probability of one variable or event, obtained from a row total or column total rather than a single interior cell.

A hiking-preference table contains 45 women, 34 coastline-preferring hikers, and 18 people who are both women and coastline-preferring. Are being female and preferring the coastline independent?

$P(F\cap C)=18/100=0.18$, while $P(F)P(C)=(45/100)(34/100)=0.153$. Since these differ, the events are not independent.

In the same hiking table, 25 men prefer lakes and streams, and 41 total hikers prefer lakes and streams. Find $P(\text{male}\mid\text{lakes and streams})$.

$P(M\mid L)=25/41\approx0.610$. The denominator is 41 because the condition defines the restricted sample space.

A mouse chooses each of three doors with probability $1/3$; the capture probabilities through the doors are $1/5$, $1/4$, and $1/2$. What is the probability of being caught through door two?

Multiply the branch probabilities: $P(\text{door 2 and caught})=(1/3)(1/4)=1/12$.

Using the three-door mouse scenario, what is the overall probability that the mouse is caught?

Add the mutually exclusive door-and-caught paths: $P(\text{caught})=1/15+1/12+1/6=19/60\approx0.317$.

Using the three-door mouse scenario, what is the probability that the mouse chooses door one or door two given that it was caught?

$P(\text{doors 1 or 2}\mid\text{caught})=\frac{1/15+1/12}{19/60}=9/19\approx0.474$.

What must the probabilities in a complete contingency table sum to?

All cell probabilities must sum to $1$. The row and column marginal probabilities must also be consistent and each sum to $1.

How does a tree diagram represent conditional probability?

Each branch represents a conditional probability given the preceding branches. Multiply probabilities along a path to obtain a joint probability, then add probabilities of disjoint paths when appropriate.

Bayes' theorem

A relationship for reversing a conditional probability: $P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}$, provided $P(B)>0$.

Why can $P(A\mid B)$ not generally be substituted for $P(B\mid A)$?

The two probabilities have different denominators and answer different questions. Confusing them can produce a base-rate error, especially when one event is uncommon.

A disease affects $14.3\%$ of a population. A test is negative for $2\%$ of people with the disease, and negative for $85\%$ of the general population. What is $P(\text{disease and negative})$?

$P(B\cap N)=P(B)P(N\mid B)=(0.143)(0.02)=0.00286\approx0.0029$.

Using the disease-testing data, what is the probability that a randomly selected person has the disease or tests negative?

$P(B\cup N)=P(B)+P(N)-P(B\cap N)=0.143+0.85-0.0029=0.9901$.

Using the disease-testing data, are having the disease and receiving a negative test mutually exclusive?

No. Their intersection has probability about $0.0029$, not zero; a diseased person can receive a false-negative result.

Using the disease-testing data, are having the disease and receiving a negative test independent?

No. $P(N\mid B)=0.02$ differs greatly from $P(N)=0.85$, so knowing that the person has the disease changes the probability of a negative result.

What is the probability of a positive test given disease if the false-negative probability is $0.02$?

Positive and negative results are complementary, so $P(\text{positive}\mid B)=1-0.02=0.98$.

What is the key sample-space change when calculating $P(A\mid B)$ from data?

The denominator is not the entire sample; it is the number of observations satisfying $B$. The numerator counts observations satisfying both $A$ and $B$.