Algebra 2 · S.CP.A.3 · S.CP.B.6

Finding a conditional probability

Once you know that one thing has happened, how do you work out the new chance of another, and how can you tell whether knowing it changed anything at all?

You need

Joint and marginal probabilityAlgebra 2 Tree diagrams and organized listsMath 7

Unlocks

Bayes' ruleAlgebra 2

Where you are 3 in Conditional probability

Probability4

Compound events10

Conditional probability3

Finding a conditional probability

The law of total probability Bayes' rule

Why this exists

The general multiplication rule builds the chance of two things together out of the chance of the first and the chance of the second given the first:

P(A \text{ and } B) = P(A) \times P(B \mid A)

Often the information comes the other way round. A bakery knows what share of its orders include a cake, and what share include a cake and are delivered; it wants to know how likely an order with a cake is to be delivered. A garden center has a two-way table of the plants it has sold and wants the chance that a fern it sold was bought as a gift. A tree diagram shows the chance of each whole path, and the chance on one branch is missing.

Each time the wanted number is a probability given an event, and the known numbers are a joint probability and the probability of that event. Dividing one by the other is the whole of this lesson. It also answers a second question: whether knowing that A happened changes the chance of B at all, which is what it means for two events to be dependent.

Definition

The conditional probability of B given A, written P(B \mid A), is the probability that B happens when it is known that A has happened. The bar is read “given”, and the event after the bar is the one known to have happened. Knowing A shrinks the possibilities to the A cases only, and P(B \mid A) is the share of those cases in which B also happens.

As a formula, for any events A and B with P(A) > 0,

P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)}

The top is the joint probability of the two events, the chance that both happen. The bottom is the probability of the event being conditioned on.

Two events A and B, each with a positive probability, are independent when knowing that one has happened does not change the chance of the other: P(B \mid A) = P(B). Otherwise they are dependent.

Intuition

At a bakery, write C for “an order includes a cake” and D for “an order is delivered”. Figure 1 shows all orders as a rectangle, with the chance of each region written in it.

Figure 1Given that an order includes a cake, only the shaded circle counts. The overlap is 0.12 of a circle worth 0.30, so P(D \mid C) = 0.4.

With nothing known, the delivered orders are the two regions inside circle D, so P(D) = 0.12 + 0.08 = 0.20. Now learn that the order includes a cake. Everything outside circle C is ruled out, and circle C, worth 0.30, is the whole world. The delivered orders left in it are the overlap, worth 0.12. The question “what share of this world is delivered?” has the answer

P(D \mid C) = \frac{0.12}{0.30} = 0.4

Dividing by 0.30 rescales the shaded circle so that it counts as 1, which is what a world has to count as. The overlap keeps its size relative to the circle. The same picture says why the formula divides by P(A) and not by anything else: A is the new whole.

Knowing C raised the chance of delivery from 0.2 to 0.4, so C and D are dependent.

Results

The formula. For any events A and B with P(A) > 0,

P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)}

From a table of counts. When outcomes are equally likely and counted in a table, the chance of B given A is the count in both, over the count in A:

P(B \mid A) = \frac{\text{number in both}}{\text{number in } A}

In a two-way table this is a cell divided by its row total, or by its column total, whichever holds the condition. The grand total cancels and is never needed.

From a tree. A second-level branch already carries a conditional probability: the branch from A to B is P(B \mid A). When it is missing, it is the path probability divided by the first-level branch, which is the formula with P(A \text{ and } B) read off the end of the path.

The complement, given A. Given A, either B happens or it does not, so

P(\text{not } B \mid A) = 1 - P(B \mid A)

Three equal tests for independence. When P(A) > 0 and P(B) > 0, each of these statements holds exactly when the other two do:

\begin{aligned} P(B \mid A) &= P(B) \\ P(A \mid B) &= P(A) \\ P(A \text{ and } B) &= P(A) \times P(B) \end{aligned}

The third is the multiplication rule for independent events. Independence is a property of the pair, so it does not matter which test is used, or which event is put first.

Why it is true

Take equally likely outcomes, which is where every probability from a deck, a die or a table of counts comes from. Say there are n outcomes, a of them in A, and k of those also in B. Knowing A leaves a outcomes, still equally likely, and k of them are in B, so the chance of B given A is \displaystyle \frac{k}{a}. Dividing top and bottom by n changes nothing:

\frac{k}{a} = \frac{k / n}{a / n} = \frac{P(A \text{ and } B)}{P(A)}

When probabilities are long-run shares rather than counts, the same argument runs with shares. In fuller treatments of probability the formula is taken as the definition of P(B \mid A), and the general multiplication rule is the same statement multiplied through by P(A).

The three tests for independence are the same equation rearranged. If P(B \mid A) = P(B), the multiplication rule gives P(A \text{ and } B) = P(A) \times P(B \mid A), which is P(A) \times P(B). If P(A \text{ and } B) = P(A) \times P(B), dividing by P(B), which is allowed because P(B) > 0, gives

P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)} = P(A)

and dividing by P(A) instead gives P(B \mid A) = P(B) in the same way. \blacksquare

Worked examples

The lesson is taught in two parts. Examples 1 to 4 belong to the first: the formula itself, and reading a conditional probability from a list of outcomes and from a two-way table. Examples 5 to 8 belong to the second: conditional probability on a tree diagram, and deciding whether two events are independent.

Example 1

From a joint probability

At a bakery, 30\% of orders include a cake and 12\% include a cake and are delivered. Also 20\% of all orders are delivered. An order is chosen at random. Find the probability that it is delivered given that it includes a cake, and the probability that it includes a cake given that it is delivered.

Solution

Write C for “includes a cake” and D for “delivered”. The percentages are shares of all orders, so P(C) = 0.30, P(C \text{ and } D) = 0.12 and P(D) = 0.20. The first question has C after the bar, so it divides by P(C):

P(D \mid C) = \frac{0.12}{0.30} = 0.4

The second has D after the bar, so it divides by P(D). The top is the same, since “cake and delivered” is the same event as “delivered and cake”:

P(C \mid D) = \frac{0.12}{0.20} = 0.6

Check in counts. Out of every 1000 orders, 300 include a cake and 120 of those are delivered, and \displaystyle \frac{120}{300} = 0.4. Of the 200 delivered orders the same 120 include a cake, and \displaystyle \frac{120}{200} = 0.6. Swapping the events kept the top and changed the bottom, so the two answers differ.

Example 2

Shrinking the list of outcomes

Two fair dice are rolled. Find the probability that they show a double (the same number on both) given that the total is 8. Then find the probability that the total is 8 given that they show a double.

Solution

There are 36 equally likely outcomes. Knowing the total is 8 leaves only these:

(2, 6),\ (3, 5),\ (4, 4),\ (5, 3),\ (6, 2)

They are still equally likely, and exactly one of the five, (4, 4), is a double. So \displaystyle P(\text{double} \mid \text{total } 8) = \frac{1}{5}. The formula agrees: one outcome of the 36 is both, and five have total 8,

\frac{1/36}{5/36} = \frac{1}{5}

For the second question the condition is now “a double”, which leaves the 6 outcomes (1, 1) to (6, 6). Again one of them, (4, 4), has total 8, so \displaystyle P(\text{total } 8 \mid \text{double}) = \frac{1}{6}. The overlap is the same single outcome both times; what changed is the world it is a share of.

Example 3

From a two-way table

A garden center records 300 plants sold by kind and by whether they were bought as a gift.

	Gift	Not a gift	Total
Rose	72	48	120
Fern	126	24	150
Cactus	18	12	30
Total	216	84	300

One sale is chosen at random. Find the probability that the plant was a gift given that it is a fern, and the probability that it is a fern given that it was a gift.

Solution

“Given that it is a fern” keeps the Fern row only: 150 plants, all equally likely to be the one chosen. Of them 126 were gifts, so

\begin{aligned} &P(\text{gift} \mid \text{fern}) \\ &\quad = \frac{126}{150} = 0.84 \end{aligned}

“Given that it was a gift” keeps the Gift column only: 216 plants, and 126 of them are ferns, so

\begin{aligned} &P(\text{fern} \mid \text{gift}) \\ &\quad = \frac{126}{216} = \frac{7}{12} \end{aligned}

which is about 0.583.

Check with the formula. \displaystyle P(\text{fern and gift}) = \frac{126}{300} = 0.42 and \displaystyle P(\text{fern}) = \frac{150}{300} = 0.5, so the formula gives \displaystyle \frac{0.42}{0.5} = 0.84, the same as the row. Dividing every count by 300 and then dividing one share by another is the same as dividing the counts, which is why the grand total never mattered.

Example 4

A condition that covers two rows

From the garden center table in Example 3, find the probability that a plant was not a gift given that it is not a fern. Then find the probability that it is a cactus given that it was not a gift.

Solution

“Not a fern” keeps the Rose and Cactus rows together, 120 + 30 = 150 plants. Of them, 48 + 12 = 60 were not gifts, so

\begin{aligned} &P(\text{not a gift} \mid \text{not a fern}) \\ &\quad = \frac{60}{150} = 0.4 \end{aligned}

“Not a gift” keeps that column, 84 plants, and 12 of them are cacti:

\begin{aligned} &P(\text{cactus} \mid \text{not a gift}) \\ &\quad = \frac{12}{84} = \frac{1}{7} \end{aligned}

A condition does not have to be a single row or column of the table. Whatever it says, first find the group it leaves, then take the share of that group.

Example 5

A missing branch on a tree

A delivery van takes the highway on 70\% of its runs and back roads on the rest. The van’s records show that 56\% of all runs take the highway and arrive on time, and 18\% take the back roads and arrive on time. Find the probability of arriving on time given each route, the probability of being late given the highway, and the probability of arriving on time on a run chosen at random.

Solution

Draw the tree with the route first. Its first branches are P(\text{highway}) = 0.7 and P(\text{back roads}) = 1 - 0.7 = 0.3. The two numbers from the records are path probabilities, the chance of a route and an arrival together. Each missing branch is its path divided by the branch it starts from:

\begin{aligned} P(\text{on time} \mid \text{highway}) &= \frac{0.56}{0.7} = 0.8 \\ P(\text{on time} \mid \text{back roads}) &= \frac{0.18}{0.3} = 0.6 \end{aligned}

Given the highway, the van is on time or late, so P(\text{late} \mid \text{highway}) = 1 - 0.8 = 0.2. A run is on time along one of two paths, and a run cannot take both routes, so

P(\text{on time}) = 0.56 + 0.18 = 0.74

Check. The highway fan adds to 0.8 + 0.2 = 1, as every fan of branches must. The formula also answers the question the other way round, “given that the van was on time, how likely is it that it took the highway?”, with \displaystyle \frac{0.56}{0.74}. Turning a tree round like that has its own lesson, Bayes’ rule.

Example 6

A condition that cuts across the paths

A box holds 10 fuses, and 3 of them are blown. Two fuses are taken at random, one after the other, without putting the first back. Find the probability that the second is blown given that the first is blown. Then find the probability that both are blown given that at least one of them is blown.

Solution

Figure 2 is the tree. Given that the first fuse is blown, the box holds 9 fuses and 2 of them are blown, so the branch from “blown” to “blown” is

P(\text{2nd blown} \mid \text{1st blown}) = \frac{2}{9}

That question needed no division at all: a second-level branch is a conditional probability already.

Figure 2Every second-level branch is already a conditional probability. A condition that cuts across the paths, such as “at least one blown”, is handled by adding the paths it keeps.

The second condition, “at least one is blown”, is not a branch. It keeps three of the four paths and rules out only good then good. Multiplying along each path gives the four path probabilities, \displaystyle \frac{6}{90}, \displaystyle \frac{21}{90}, \displaystyle \frac{21}{90} and \displaystyle \frac{42}{90}, which add to 1. So

\begin{aligned} P(\text{at least one}) &= 1 - \frac{42}{90} = \frac{48}{90} \\ P(\text{both blown}) &= \frac{6}{90} \end{aligned}

“Both blown and at least one blown” is just “both blown”, since both blown is one of the kept paths. The formula gives

\begin{aligned} &P(\text{both} \mid \text{at least one}) \\ &\quad = \frac{6/90}{48/90} = \frac{6}{48} = \frac{1}{8} \end{aligned}

Knowing that some fuse is blown makes “both blown” more likely than it was, \displaystyle \frac{1}{8} against \displaystyle \frac{6}{90} = \frac{1}{15}, but far from certain.

Example 7

Testing independence from a table

A camera club of 200 members is recorded by the camera each member mostly uses and the subject each mostly photographs.

	Landscape	Portrait	Total
Film	12	18	30
Digital	68	102	170
Total	80	120	200

A member is chosen at random. Are “uses film” and “photographs landscapes” independent? Compare “fern” and “gift” at the garden center of Example 3.

Solution

With nothing known, \displaystyle P(\text{landscape}) = \frac{80}{200} = 0.4. Given film, the Film row of 30 has 12 landscape members:

P(\text{landscape} \mid \text{film}) = \frac{12}{30} = 0.4

The two are equal, so knowing the camera does not change the chance of landscapes, and the events are independent. Either of the other tests says the same. By the product, \displaystyle P(\text{film}) = \frac{30}{200} = 0.15 and

\begin{aligned} &P(\text{film}) \times P(\text{landscape}) \\ &\quad = 0.15 \times 0.4 = 0.06 \end{aligned}

which equals \displaystyle P(\text{film and landscape}) = \frac{12}{200} = 0.06. And with the condition the other way round, \displaystyle P(\text{film} \mid \text{landscape}) = \frac{12}{80} = 0.15, which is P(\text{film}) again.

At the garden center, \displaystyle P(\text{gift}) = \frac{216}{300} = 0.72 but P(\text{gift} \mid \text{fern}) = 0.84. Knowing the plant is a fern raises the chance, so those events are dependent.

Example 8

Independence from the addition rule

At a library, 60\% of visitors borrow a book, 25\% use a computer, and 70\% do at least one of the two. A visitor is chosen at random. Are “borrows a book” and “uses a computer” independent? Find the probability that a visitor who uses a computer borrows a book.

Solution

Write B for borrows a book and U for uses a computer. The formula needs the joint probability, which is not given. The general addition rule, P(B \text{ or } U) = P(B) + P(U) - P(B \text{ and } U), can be solved for it:

\begin{aligned} P(B \text{ and } U) &= 0.6 + 0.25 - 0.7 \\ &= 0.15 \end{aligned}

Then

P(B \mid U) = \frac{0.15}{0.25} = 0.6

and P(B) = 0.6 as well, so using a computer does not change the chance of borrowing: the events are independent. The product test agrees, 0.6 \times 0.25 = 0.15. A visitor who uses a computer borrows a book with probability 0.6, the same as any visitor.

Boundaries

The condition must be possible. P(B \mid A) is defined only when P(A) > 0. The formula would divide by zero otherwise, and there would be no A cases to take a share of.

Which event is after the bar matters. P(B \mid A) and P(A \mid B) share a top and differ in the bottom, so they are equal only when P(A) = P(B) (or the events cannot happen together). In Example 3, 0.84 of the ferns were gifts but only \displaystyle \frac{7}{12} of the gifts were ferns. A sentence like “40\% of cake orders are delivered” fixes the condition; “40\% of orders include a cake and are delivered” is a joint probability, not a conditional one.

The top is the chance of both, not a product. P(A \text{ and } B) may be replaced by P(A) \times P(B) only when A and B are already known to be independent, and then the formula just returns P(B).

Independence is a statement about exact probabilities. The tests above decide independence when the probabilities themselves are known, as with dice, a deck, or a population counted in full. Shares in a sample from a larger group wobble from sample to sample, so two shares that are close, or even equal, in a sample show only that the events may be independent in the whole group. Deciding that from data is a question for statistics, not settled by this formula.

Independent is not the same as mutually exclusive. If A and B cannot happen together and both have positive probabilities, then P(B \mid A) = 0, which is not P(B). Such events are always dependent: knowing A rules B out completely.

Where this goes next

The formula turns any probability “given A” into two probabilities about the whole population, and that is what lets a condition be turned round. Knowing P(B \mid A) for every branch of a tree, the chance of the reverse, P(A \mid B), is the path through A divided by the total of all paths that end in B. That is Bayes’ rule, and the total in its bottom comes from the law of total probability. Both belong to the wider study of conditional probability.

Independence, tested here one pair at a time, is what justifies multiplying chances in repeated independent trials, and the conditional shares of a table are the probability version of the conditional relative frequencies used to look for an association between two variables in data.

Exercises

Exercises 1 to 10 go with the first part of the lesson (Examples 1 to 4), Exercises 11 to 20 with the second (Examples 5 to 8), and the reasoning group draws on both.

Part 1: the formula

P(A \text{ and } B) = 0.18 and P(A) = 0.6. Find P(B \mid A).
P(A \text{ and } B) = 0.1 and P(B) = 0.25. Find P(A \mid B).
At a beach, 80\% of summer days are sunny, and on 50\% of summer days it is sunny and the parking lot fills up. Find the probability that the parking lot fills up on a summer day given that it is sunny.
At a castle, G is “a visitor goes to the gift shop” and K is “a visitor goes to the cafe”. Write each statement in symbols: (a) 40\% of the cafe’s visitors also go to the gift shop; (b) a quarter of all visitors go to both; (c) of those who go to the gift shop, 60\% go to the cafe. Then use the three statements to find P(K) and P(G).
One fair die is rolled. Find the probability that the result is greater than 3 given that it is even, and the probability that it is a prime number given that it is greater than 3.
Three fair coins are tossed. Find the probability that all three land heads given that at least two land heads.

Part 1: two-way tables

An evening pottery class has 150 students, recorded by the method each mostly uses and by level.

	Beginner	Advanced	Total
Wheel	54	36	90
Hand-building	30	30	60
Total	84	66	150

Find P(\text{advanced} \mid \text{wheel}) and P(\text{wheel} \mid \text{advanced}).
Find P(\text{hand-building} \mid \text{beginner}) and P(\text{beginner} \mid \text{hand-building}).
Find P(\text{wheel and beginner}) and P(\text{wheel}) as shares of all 150 students, and use the formula to find P(\text{beginner} \mid \text{wheel}). Check it against the Wheel row.
Of 400 people surveyed, 160 are under 30. Of those under 30, 35\% walk to work; of those aged 30 or over, 15\% walk to work. Build the two-way table, then find the probability that a person chosen at random walks to work, and the probability that a person is under 30 given that they walk to work.

Part 2: trees

On a tree, the first branch to A carries 0.4, the path to “A and B” ends in 0.1, and the path to “not A and B” ends in 0.3. Find P(B \mid A), P(B \mid \text{not } A) and P(B). Are A and B independent?
A tin holds 6 plain and 4 chocolate cookies. Two are taken at random without replacement. Find the probability that the second is plain given that the first is plain, and the probability that both are plain given that at least one is plain.
For the tin in Exercise 12, find the probability that both cookies are plain given that the two are the same kind.
Two fair dice are rolled. Find the probability that both show a number greater than 4 given that at least one does.

Part 2: independent or not

P(A) = 0.6, P(B) = 0.35 and P(A \text{ and } B) = 0.21. Find P(B \mid A) and decide whether A and B are independent.
P(A) = 0.5, P(B) = 0.3 and P(A \text{ and } B) = 0.2. Find P(B \mid A) and P(A \mid B), and decide whether A and B are independent.
An art gallery counts 250 visitors: 100 buy a postcard, 100 come at the weekend, and 40 buy a postcard at the weekend. Is “buys a postcard” independent of “comes at the weekend”?
One fair die is rolled. E is “even”, F is “at most 4” and H is “at most 3”. Is E independent of F? Is E independent of H?
P(A) = 0.4, P(B) = 0.5 and P(A \text{ or } B) = 0.8. Find P(B \mid A) and decide whether A and B are independent.
P(A) = 0.75 and P(A \text{ and } B) = 0.3. What must P(B) be for A and B to be independent?

Reasoning

Explain why P(B \mid A) and P(A \mid B) are usually different, and say when they are equal.
Show that if P(A) > 0, P(B) > 0 and P(B \mid A) = P(B), then P(A \mid B) = P(A).
Explain why P(A \mid A) = 1 and why P(\text{not } B \mid A) = 1 - P(B \mid A) whenever P(A) > 0.
Explain why two mutually exclusive events, each with a positive probability, can never be independent.
Show that if A and B are independent, then A and “not B” are independent too.

Answers

0.3. \displaystyle \frac{0.18}{0.6} = 0.3.
0.4. \displaystyle \frac{0.1}{0.25} = 0.4; the condition is B, so divide by P(B).
0.625. Divide by the chance of the condition, sunny: \displaystyle \frac{0.5}{0.8} = 0.625.
(a) P(G \mid K) = 0.4; (b) P(G \text{ and } K) = 0.25; (c) P(K \mid G) = 0.6. From (a) and (b), \displaystyle 0.4 = \frac{0.25}{P(K)}, so P(K) = 0.625. From (b) and (c), \displaystyle 0.6 = \frac{0.25}{P(G)}, so \displaystyle P(G) = \frac{5}{12}, about 0.417.
\displaystyle \frac{2}{3} and \displaystyle \frac{1}{3}. Even leaves 2, 4, 6, and two of them are greater than 3. Greater than 3 leaves 4, 5, 6, and only 5 is prime.
\displaystyle \frac{1}{4}. At least two heads leaves HHT, HTH, THH and HHH, equally likely, and one of them is all heads.
0.4 and \displaystyle \frac{6}{11}. The Wheel row gives \displaystyle \frac{36}{90} = 0.4; the Advanced column gives \displaystyle \frac{36}{66} = \frac{6}{11}.
\displaystyle \frac{5}{14} and 0.5. The Beginner column gives \displaystyle \frac{30}{84} = \frac{5}{14}; the Hand-building row gives \displaystyle \frac{30}{60} = 0.5.
0.36, 0.6 and 0.6. \displaystyle \frac{54}{150} = 0.36 and \displaystyle \frac{90}{150} = 0.6, so the formula gives \displaystyle \frac{0.36}{0.6} = 0.6, and the row gives \displaystyle \frac{54}{90} = 0.6 as well.
0.23 and \displaystyle \frac{14}{23}, about 0.609. Under 30: 0.35 \times 160 = 56 walk and 104 do not. Aged 30 or over, 240 people: 0.15 \times 240 = 36 walk and 204 do not. So 92 of the 400 walk, \displaystyle \frac{92}{400} = 0.23, and of those 92, 56 are under 30: \displaystyle \frac{56}{92} = \frac{14}{23}.
P(B \mid A) = 0.25, P(B \mid \text{not } A) = 0.5, P(B) = 0.4; dependent. \displaystyle \frac{0.1}{0.4} and \displaystyle \frac{0.3}{0.6}, since the branch to not A is 0.6; the two paths to B add to 0.4, which is not 0.25.
\displaystyle \frac{5}{9} and \displaystyle \frac{5}{13}. After a plain one, 5 of the 9 left are plain. Both plain is \displaystyle \frac{6}{10} \times \frac{5}{9} = \frac{1}{3}; none plain is \displaystyle \frac{4}{10} \times \frac{3}{9} = \frac{2}{15}, so at least one plain is \displaystyle \frac{13}{15}, and \displaystyle \frac{1/3}{13/15} = \frac{5}{13}.
\displaystyle \frac{5}{7}. Same kind is both plain or both chocolate, \displaystyle \frac{1}{3} + \frac{2}{15} = \frac{7}{15}, and \displaystyle \frac{1/3}{7/15} = \frac{5}{7}.
\displaystyle \frac{1}{5}. Of the 36 outcomes, 4 \times 4 = 16 have neither die above 4, so 20 have at least one; 2 \times 2 = 4 have both, and \displaystyle \frac{4}{20} = \frac{1}{5}.
P(B \mid A) = 0.35; independent. \displaystyle \frac{0.21}{0.6} = 0.35, which equals P(B).
P(B \mid A) = 0.4 and \displaystyle P(A \mid B) = \frac{2}{3}; dependent. \displaystyle \frac{0.2}{0.5} = 0.4 \ne 0.3, and \displaystyle \frac{0.2}{0.3} = \frac{2}{3} \ne 0.5.
Yes, independent. \displaystyle P(\text{weekend} \mid \text{postcard}) = \frac{40}{100} = 0.4 and \displaystyle P(\text{weekend}) = \frac{100}{250} = 0.4.
E and F are independent; E and H are dependent. \displaystyle P(E) = \frac{1}{2}. F leaves 1, 2, 3, 4, two of them even, so \displaystyle P(E \mid F) = \frac{1}{2}. H leaves 1, 2, 3, one of them even, so \displaystyle P(E \mid H) = \frac{1}{3}.
P(B \mid A) = 0.25; dependent. By the addition rule P(A \text{ and } B) = 0.4 + 0.5 - 0.8 = 0.1, and \displaystyle \frac{0.1}{0.4} = 0.25 \ne 0.5.
P(B) = 0.4. Independence needs \displaystyle P(B) = P(B \mid A) = \frac{0.3}{0.75} = 0.4.
Both have P(A \text{ and } B) on top, but one divides by P(A) and the other by P(B): each is a share of a different group. They are equal exactly when P(A) = P(B), or when P(A \text{ and } B) = 0 so that both are 0.
P(B \mid A) = P(B) means \displaystyle \frac{P(A \text{ and } B)}{P(A)} = P(B), so P(A \text{ and } B) = P(A) \times P(B). Dividing by P(B) > 0 gives \displaystyle \frac{P(A \text{ and } B)}{P(B)} = P(A), which is P(A \mid B) = P(A).
“A and A” is just A, so \displaystyle P(A \mid A) = \frac{P(A)}{P(A)} = 1: once A is known, it is certain. Given A, every A outcome is in B or not in B and not both, so the two shares of the A cases add to 1.
Mutually exclusive means P(A \text{ and } B) = 0, so \displaystyle P(B \mid A) = \frac{0}{P(A)} = 0. But P(B) > 0, so P(B \mid A) \ne P(B). Knowing A happened rules B out, which is the opposite of changing nothing.
A splits into “A and B” and “A and not B”, which cannot both happen, so P(A \text{ and not } B) = P(A) - P(A \text{ and } B). By independence this is P(A) - P(A) \times P(B), which factors as P(A) \times (1 - P(B)), that is P(A) \times P(\text{not } B). This is the product test for A and not B.