Conditional probability

Conditional probability is the probability of an event when some information is already known. If $B$ is known to have occurred, then the possible outcomes are restricted to $B$. The conditional probability of $A$ given $B$ is

$$\boxed{\operatorname{P}(A\mid B)=\frac{\operatorname{P}(A\cap B)}{\operatorname{P}(B)}} \qquad \text{provided }\operatorname{P}(B)>0.$$

The symbol $A\mid B$ is read as ”$A$ given $B$”. It does not mean division. The event after the vertical bar is the information assumed to be true.

Prerequisites

You should be able to:

• use complements, intersections and unions from probability;
• distinguish mutually exclusive events from independent events;
• multiply along branches and add between alternative routes in a probability tree;
• form fractions from frequencies in a table.

Restricting the sample space

Suppose a fair die is rolled. Let

$$A=\{\text{the result is even}\},\qquad B=\{\text{the result is greater than }3\}.$$

Without extra information, $A=\{2,4,6\}$ and $\operatorname{P}(A)=3/6$. If $B$ is known, the relevant sample space shrinks from $\{1,2,3,4,5,6\}$ to $\{4,5,6\}$. Two of these three outcomes are even, so

$$\operatorname{P}(A\mid B)=\frac23.$$

The formula gives the same result:

$$\operatorname{P}(A\mid B) =\frac{\operatorname{P}(A\cap B)}{\operatorname{P}(B)} =\frac{2/6}{3/6} =\frac23.$$

The denominator is $\operatorname{P}(B)$ because knowing $B$ has occurred makes $B$ the new whole.

The order matters

In general,

$$\operatorname{P}(A\mid B)\ne \operatorname{P}(B\mid A).$$

The event after the bar determines the restricted sample space. Always translate the notation into words before calculating.

Worked example 1: cards without replacement

One card is chosen at random from an ordinary pack of $52$ cards. Given that it is a picture card, find the probability that it is a king.

There are $12$ picture cards: four jacks, four queens and four kings. Once “picture card” is known, only these $12$ cards remain relevant. Therefore

$$\boxed{\operatorname{P}(\text{king}\mid\text{picture})=\frac4{12}=\frac13.}$$

Using all $52$ cards in the denominator would ignore the given information.

Self-check 1

A student is selected from a group of $18$ students. Ten study physics, seven study chemistry, and four study both. Given that the student studies physics, find the probability that the student also studies chemistry.

Answer

The new whole is the $10$ physics students. Four of them also study chemistry, so

$$\boxed{\operatorname{P}(C\mid P)=\frac4{10}=\frac25.}$$

The multiplication rule

Rearranging the conditional probability formula gives

$$\boxed{\operatorname{P}(A\cap B)=\operatorname{P}(B)\operatorname{P}(A\mid B).}$$

Equally, reversing the roles of the events gives

$$\boxed{\operatorname{P}(A\cap B)=\operatorname{P}(A)\operatorname{P}(B\mid A).}$$

These equations describe the same intersection by entering it from different directions. They also explain the multiply rule on a tree diagram: the probability at the end of a route is the product of the branch probabilities along that route.

Worked example 2: dependent selections

A bag contains $5$ red and $3$ blue counters. Two counters are selected without replacement. Find the probability that both are red.

After a red counter is selected first, $4$ red counters remain among $7$ counters. Thus

$$\operatorname{P}(R_1)=\frac58, \qquad \operatorname{P}(R_2\mid R_1)=\frac47.$$

Hence

$$\begin{aligned} \operatorname{P}(R_1\cap R_2) &=\operatorname{P}(R_1)\operatorname{P}(R_2\mid R_1)\\ &=\frac58\times\frac47\\ &=\boxed{\frac5{14}}. \end{aligned}$$

The second probability is conditional. Without replacement, the contents of the bag change, so writing $5/8\times5/8$ would be incorrect.

Worked example 3: at least one success

The same bag contains $5$ red and $3$ blue counters. Two are selected without replacement. Find the probability of at least one red counter.

The complement of “at least one red” is “both blue”. Therefore

$$\begin{aligned} \operatorname{P}(\text{at least one red}) &=1-\operatorname{P}(B_1\cap B_2)\\ &=1-\frac38\times\frac27\\ &=1-\frac3{28}\\ &=\boxed{\frac{25}{28}}. \end{aligned}$$

Using a complement avoids separately calculating the routes red then red, red then blue, and blue then red.

Tree diagrams with conditional probabilities

On a tree diagram:

• branches leaving the same point must sum to $1$;
• multiply probabilities along one complete route;
• add probabilities for mutually exclusive routes that satisfy the question.

The second set of branches may depend on the outcome of the first event. Labels should make this dependence explicit.

Worked example 4: combining routes

A factory uses two machines. Machine $M_1$ makes $60\%$ of the components and machine $M_2$ makes the rest. Of the components made by $M_1$, $2\%$ are defective. Of those made by $M_2$, $5\%$ are defective. Find the probability that a randomly selected component is defective.

There are two mutually exclusive routes to a defective component:

$$M_1\cap D \qquad\text{or}\qquad M_2\cap D.$$

Calculate each route, then add:

$$\begin{aligned} \operatorname{P}(D) &=\operatorname{P}(M_1)\operatorname{P}(D\mid M_1) +\operatorname{P}(M_2)\operatorname{P}(D\mid M_2)\\ &=(0.60)(0.02)+(0.40)(0.05)\\ &=0.012+0.020\\ &=\boxed{0.032}. \end{aligned}$$

This is an instance of the law of total probability: split an event into all possible routes leading to it.

Worked example 5: reversing the condition

For the factory in Example 4, a selected component is known to be defective. Find the probability that it was made by $M_2$.

The question asks for $\operatorname{P}(M_2\mid D)$, not $\operatorname{P}(D\mid M_2)$. Use the route probability from the tree and divide by the total probability of being defective:

$$\begin{aligned} \operatorname{P}(M_2\mid D) &=\frac{\operatorname{P}(M_2\cap D)}{\operatorname{P}(D)}\\ &=\frac{(0.40)(0.05)}{0.032}\\ &=\frac{0.020}{0.032}\\ &=\boxed{0.625}. \end{aligned}$$

Although $M_2$ makes only $40\%$ of all components, it accounts for $62.5\%$ of defective components because its defect rate is higher.

A reliable reverse probability method

1. Identify exactly what is after the bar.
2. Find the probability of the required joint route for the numerator.
3. Add every route consistent with the given information for the denominator.
4. Divide only after both quantities are known.

Self-check 2

A disease affects $1\%$ of a population. A test is positive for $95\%$ of people with the disease and for $4\%$ of people without it. A person tests positive. Find the probability that they have the disease.

Answer

Let $D$ mean the person has the disease and $+$ mean a positive test. First find the two routes to a positive result:

$$\operatorname{P}(D\cap +)=(0.01)(0.95)=0.0095,$$ $$\operatorname{P}(D'\cap +)=(0.99)(0.04)=0.0396.$$

Therefore

$$\begin{aligned} \operatorname{P}(D\mid +) &=\frac{0.0095}{0.0095+0.0396}\\ &=\frac{0.0095}{0.0491}\\ &\approx\boxed{0.193}. \end{aligned}$$

The result is much smaller than $0.95$. The $95\%$ is $\operatorname{P}(+\mid D)$, which is the reverse conditional probability.

Conditional probability from a two way table

Tables are particularly useful when data are given as frequencies. Restrict attention to the row or column representing the condition, then form a proportion within that total.

Worked example 6: reading both directions

A survey records how $200$ people travel to work.

	Car	Public transport	Total
Under 40	54	66	120
40 or over	48	32	80
Total	102	98	200

Let $C$ mean travels by car and $U$ mean under 40.

Given that a person is under 40, the relevant total is $120$:

$$\operatorname{P}(C\mid U)=\frac{54}{120}=\boxed{0.45}.$$

Given that a person travels by car, the relevant total is $102$:

$$\operatorname{P}(U\mid C)=\frac{54}{102}=\boxed{\frac9{17}}.$$

Both calculations use the same intersection frequency, $54$, but their denominators differ because the conditions differ.

Frequencies and probabilities

If every entry in a table is a frequency, you can calculate directly with frequencies. The common grand total would cancel:

$$\frac{54/200}{120/200}=\frac{54}{120}.$$

Self-check 3

Using the table above, find:

1. $\operatorname{P}(U\mid C')$;
2. $\operatorname{P}(C'\mid U')$.

Answer

1. $C'$ means public transport. Of the $98$ public transport users, $66$ are under 40:

   $$\boxed{\operatorname{P}(U\mid C')=\frac{66}{98}=\frac{33}{49}}.$$

2. $U'$ means aged 40 or over. Of these $80$ people, $32$ use public transport:

   $$\boxed{\operatorname{P}(C'\mid U')=\frac{32}{80}=\frac25}.$$

Independence

Events $A$ and $B$ are independent if knowing that one has occurred does not change the probability of the other. Provided the relevant conditional probability is defined,

$$\boxed{\operatorname{P}(A\mid B)=\operatorname{P}(A).}$$

Using the multiplication rule, this is equivalent to

$$\boxed{\operatorname{P}(A\cap B)=\operatorname{P}(A)\operatorname{P}(B).}$$

Either statement can be used to test independence. The product test is often most convenient when joint and marginal probabilities are given.

Worked example 7: testing independence

For the travel survey in Example 6, determine whether being under 40 and travelling by car are independent.

From the table,

$$\operatorname{P}(C)=\frac{102}{200}=0.51$$

but

$$\operatorname{P}(C\mid U)=\frac{54}{120}=0.45.$$

Since $0.45\ne0.51$, knowing that a person is under 40 changes the probability that they travel by car. Therefore $C$ and $U$ are not independent.

We could instead compare

$$\operatorname{P}(C\cap U)=\frac{54}{200}=0.27$$

with

$$\operatorname{P}(C)\operatorname{P}(U) =\frac{102}{200}\times\frac{120}{200} =0.306.$$

These are unequal, giving the same conclusion.

Worked example 8: finding an unknown using independence

Events $A$ and $B$ are independent, with

$$\operatorname{P}(A)=0.4, \qquad \operatorname{P}(A\cup B)=0.7.$$

Find $\operatorname{P}(B)$.

Let $\operatorname{P}(B)=p$. Independence gives

$$\operatorname{P}(A\cap B)=0.4p.$$

Use the addition rule:

$$\begin{aligned} \operatorname{P}(A\cup B) &=\operatorname{P}(A)+\operatorname{P}(B)-\operatorname{P}(A\cap B)\\ 0.7&=0.4+p-0.4p\\ 0.3&=0.6p\\ p&=0.5. \end{aligned}$$

Therefore

$$\boxed{\operatorname{P}(B)=0.5.}$$

Independent does not mean mutually exclusive

If nonzero probability events are mutually exclusive, then $\operatorname{P}(A\cap B)=0$. If they were independent, the intersection would instead have probability $\operatorname{P}(A)\operatorname{P}(B)>0$. Such events cannot be both mutually exclusive and independent.

Self-check 4

Events $A$ and $B$ satisfy

$$\operatorname{P}(A)=0.3,\qquad \operatorname{P}(B)=0.6,\qquad \operatorname{P}(A\cap B)=0.18.$$

Are $A$ and $B$ independent? Find $\operatorname{P}(A\mid B)$.

Answer

Since

$$\operatorname{P}(A)\operatorname{P}(B)=(0.3)(0.6)=0.18 =\operatorname{P}(A\cap B),$$

the events are independent. Also,

$$\operatorname{P}(A\mid B)=\frac{0.18}{0.6}=\boxed{0.3},$$

which equals $\operatorname{P}(A)$, as expected.

Solving algebraic probability problems

Conditional probability can determine an unknown parameter. Keep probabilities exact where possible and check that every solution lies between $0$ and $1$.

Worked example 9: using a given condition

Suppose

$$\operatorname{P}(A)=0.55,\qquad \operatorname{P}(B)=0.40,\qquad \operatorname{P}(A\cup B)=0.70.$$

Find $\operatorname{P}(A\mid B)$.

First use the addition rule to find the intersection:

$$\begin{aligned} \operatorname{P}(A\cap B) &=\operatorname{P}(A)+\operatorname{P}(B)-\operatorname{P}(A\cup B)\\ &=0.55+0.40-0.70\\ &=0.25. \end{aligned}$$

Now condition on $B$:

$$\operatorname{P}(A\mid B) =\frac{0.25}{0.40} =\boxed{0.625}.$$

Worked example 10: reconstructing a probability tree

A student either walks or takes a bus to college. The probability that the student walks is $w$. If the student walks, the probability of arriving late is $0.1$. If the student takes the bus, it is $0.3$. The overall probability of arriving late is $0.22$. Find $w$.

The probabilities of walking and taking the bus are $w$ and $1-w$. Add the two routes to being late:

$$\begin{aligned} 0.22&=0.1w+0.3(1-w)\\ 0.22&=0.1w+0.3-0.3w\\ -0.08&=-0.2w\\ w&=0.4. \end{aligned}$$

Thus

$$\boxed{\operatorname{P}(\text{walks})=0.4.}$$

As a check, $(0.4)(0.1)+(0.6)(0.3)=0.04+0.18=0.22$.

Common misconceptions

Swapping the condition

$\operatorname{P}(A\mid B)$ restricts the sample space to $B$. It is not usually equal to $\operatorname{P}(B\mid A)$. In context, write both probabilities in words.

Dividing by the wrong probability

The denominator is the probability of the event after the bar:

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

Assuming independence

Multiplying $\operatorname{P}(A)$ by $\operatorname{P}(B)$ is valid only for independent events. In general, use

$$\operatorname{P}(A\cap B)=\operatorname{P}(A)\operatorname{P}(B\mid A).$$

Adding routes that overlap

Alternative complete routes on a tree are mutually exclusive, so they can be added. For general events that overlap, use

$$\operatorname{P}(A\cup B) =\operatorname{P}(A)+\operatorname{P}(B)-\operatorname{P}(A\cap B).$$

Rounding too early

Keep route probabilities and totals exact until the final line. Early rounding can noticeably alter a reverse conditional probability because one rounded value is divided by another.

Mixed self-check

1. A box contains $6$ green and $4$ yellow balls. Two balls are selected without replacement. Find the probability that the second ball is yellow given that the first is green.
2. Given $\operatorname{P}(A)=0.7$, $\operatorname{P}(B)=0.5$ and $\operatorname{P}(A\cap B)=0.35$, determine whether $A$ and $B$ are independent.
3. A shop receives $70\%$ of its batteries from supplier $X$ and the remainder from supplier $Y$. The failure rates are $1\%$ for $X$ and $3\%$ for $Y$. Find the probability that a failed battery came from $Y$.
4. Events $A$ and $B$ satisfy $\operatorname{P}(A\mid B)=0.6$ and $\operatorname{P}(B)=0.25$. Find $\operatorname{P}(A\cap B)$.

Answers

1. After a green ball is removed, $4$ of the remaining $9$ balls are yellow:

   $$\boxed{\operatorname{P}(Y_2\mid G_1)=\frac49}.$$

2. Since

   $$\operatorname{P}(A)\operatorname{P}(B)=(0.7)(0.5)=0.35 =\operatorname{P}(A\cap B),$$

   the events are independent.

3. The failed routes have probabilities

   $$(0.70)(0.01)=0.007 \quad\text{and}\quad (0.30)(0.03)=0.009.$$

   Hence

   $$\operatorname{P}(Y\mid F) =\frac{0.009}{0.007+0.009} =\boxed{0.5625}.$$

4. Rearrange the conditional probability formula:

   $$\operatorname{P}(A\cap B) =\operatorname{P}(B)\operatorname{P}(A\mid B) =(0.25)(0.6) =\boxed{0.15}.$$

Summary

The essential ideas are:

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

and, for independent events,

$$\operatorname{P}(A\cap B) =\operatorname{P}(A)\operatorname{P}(B).$$

Conditioning changes the sample space. On trees, multiply along routes and add alternative routes. In tables, use the total belonging to the condition. Never reverse the events on either side of the bar without justification.

Next steps

• Review probability if intersections, complements or independence need more practice.
• Study probability modelling to judge whether assumptions such as independence are reasonable.
• Apply repeated independent trials in the binomial distribution.
• Use probability models with discrete random variables.