Algebra 2 · S.CP.B.8

The general multiplication rule

Two things both have to happen, and the first one changes the chances of the second. Is there one rule for the chance of both that works every time?

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The general multiplication rule

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Why this exists

When two events are independent, the multiplication rule for independent events gives the chance of both: multiply the two chances. When they are not, the lesson on dependent events showed what to do for draws from a jar or a hat: recount what is left after the first draw and multiply by the new chance.

Many pairs of events are not draws at all. A town’s commuters are counted, and 40\% of them take the train; among the train takers, 15\% cycle to the station. Nothing is taken out of anything, and yet the chance of cycling clearly depends on whether the person takes the train. Or a cable car shuts more often on windy days than on calm ones. Here there is nothing to recount, but there is still a “chance of the second, once the first is known”.

This lesson gives that chance a name and a symbol, and states one rule for the chance of both that covers every case: independent events, draws without replacement, tables of counts, and chains of three or more events.

Definition

The probability of B given A, written P(B \mid A), is the probability that B happens when it is known that A has happened. It is the share of the A cases in which B also happens. The bar is read “given”, and the event after the bar is the one known to have happened. It only makes sense when P(A) > 0, since there must be some A cases to take a share of.

In the commuter town, write T for “takes the train” and C for “cycles to the station”. Then P(T) = 0.4 and P(C \mid T) = 0.15. The 0.15 is a share of the train takers, not of the whole town.

The general multiplication rule says that for any two events A and B with P(A) > 0,

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

In words: the chance that both happen is the chance of the first, times the chance of the second given the first. “A and B” is the intersection of the two events, also written A \cap B. Its probability is the joint probability of A and B.

Intuition

The rule is a fraction of a fraction. Picture all the commuters as a square of area 1, as in Figure 1. The train takers fill a strip 0.4 of the width. Inside that strip, the cyclists fill 0.15 of its height.

Figure 1The shaded piece is 0.15 of a strip that is 0.4 of the square, so it is 0.4 \times 0.15 = 0.06 of the square.

The shaded piece is 0.4 wide and 0.15 tall, so its area is

P(T \text{ and } C) = 0.4 \times 0.15 = 0.06

Six commuters in every hundred take the train and cycle to the station. The second factor was a share of the strip, and the first factor says how big the strip is. Multiplying turns “a share of a share” into a share of the whole.

The same picture explains why the second factor must be the conditional one. The cyclists who do not take the train sit somewhere in the other strip, and they are not part of the shaded piece at all. Only the share of cyclists within the train strip tells you how much of that strip is shaded.

Results

The general multiplication rule. For any events A and B with P(A) > 0,

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

Either event can go first. “A and B” is the same event as “B and A”, so when P(B) > 0 also

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

The two products are equal even though their factors are usually different.

The independent case. A and B are independent exactly when knowing A changes nothing, that is P(B \mid A) = P(B). Then the rule becomes

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

which is the rule for independent events. So that rule is a special case of this one, not a different rule.

Three events. Each new factor is given everything before it:

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

and a fourth event adds a factor given all three earlier ones, and so on.

On a tree. In a tree diagram, a first-level branch carries P(A) and the branch that follows it carries P(B \mid A). Multiplying along a path is the general multiplication rule, applied once for each step.

Why it is true

Take the case where the outcomes are equally likely, which is where every probability in a jar, a deck or a table of counts comes from. Say there are n outcomes, a of them are in A, and k of those are in B as well. So k outcomes are in “A and B”. Then

P(A) = \frac{a}{n}, \qquad P(B \mid A) = \frac{k}{a}

because P(B \mid A) is the share of the A outcomes that are also in B. Multiplying,

P(A) \times P(B \mid A) = \frac{a}{n} \times \frac{k}{a} = \frac{k}{n}

and \displaystyle \frac{k}{n} is exactly P(A \text{ and } B): the a cancels. The rule is the statement that “the share of the whole” equals “the share of A in the whole” times “the share of B inside A”.

When probabilities are long-run shares rather than counts, as with the commuters, the same argument runs with shares in place of counts: of every 1 commuter, 0.4 take the train, and of that 0.4, a share 0.15 also cycle. In fuller treatments of probability, this is turned round and used as the definition of P(B \mid A): it is the number that makes the rule true. Dividing both sides by P(A) gives the formula for a conditional probability, which is the subject of finding a conditional probability.

For three events, apply the two-event rule twice. Treat “A and B” as one event, so that

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

then replace P(A \text{ and } B) by P(A) \times P(B \mid A).

Worked examples

Example 1

Two seed packets from a jar

A jar holds 18 seed packets, 5 of them tomato. Two packets are drawn at random, one after the other, and the first is not put back. Find the probability that both are tomato. Then find the probability that the second packet is tomato when nothing is known about the first, and compare it with the conditional probability used in the first answer.

Solution

Name the events: A is “the first packet is tomato” and B is “the second packet is tomato”. The first draw is from all 18 packets, so \displaystyle P(A) = \frac{5}{18}. Given A, the jar holds 17 packets and 4 of them are tomato, so \displaystyle P(B \mid A) = \frac{4}{17}. By the general multiplication rule,

P(A \text{ and } B) = \frac{5}{18} \times \frac{4}{17} = \frac{20}{306} = \frac{10}{153}

which is about 0.065.

With nothing known about the first packet, the second is tomato along two paths: tomato then tomato, or another kind then tomato. They cannot both happen, so their probabilities add:

\begin{aligned} P(B) &= \frac{5}{18} \times \frac{4}{17} + \frac{13}{18} \times \frac{5}{17} \\ &= \frac{20}{306} + \frac{65}{306} = \frac{85}{306} = \frac{5}{18} \end{aligned}

So \displaystyle P(B) = \frac{5}{18} \approx 0.278, while \displaystyle P(B \mid A) = \frac{4}{17} \approx 0.235. Knowing the first packet was tomato lowers the chance for the second, so A and B are dependent, and the product P(A) \times P(B) would not give the chance of both. The general rule asks for the second factor given the first, and that is \displaystyle \frac{4}{17}.

Example 2

Either order, from a table

In a 10 km race, 200 runners finished. The table counts them by whether they live locally and whether they finished in under an hour.

	Under an hour	An hour or more	Total
Local	42	78	120
Visitor	18	62	80
Total	60	140	200

A finisher is chosen at random. Let L be “local” and F be “under an hour”. Find P(L \text{ and } F) with the general multiplication rule in both orders, and check both against the table.

Solution

Local first. \displaystyle P(L) = \frac{120}{200} = 0.6. Given that the runner is local, we are inside the Local row of 120 runners, and 42 of them were under an hour, so \displaystyle P(F \mid L) = \frac{42}{120} = 0.35. Then

P(L \text{ and } F) = 0.6 \times 0.35 = 0.21

Fast first. \displaystyle P(F) = \frac{60}{200} = 0.3. Given that the runner was under an hour, we are inside that column of 60 runners, and 42 of them are local, so \displaystyle P(L \mid F) = \frac{42}{60} = 0.7. Then

P(F \text{ and } L) = 0.3 \times 0.7 = 0.21

Check. The cell for local and under an hour holds 42 of the 200 runners, and \displaystyle \frac{42}{200} = 0.21. Both orders agree with it, as they must, since both describe the same 42 people. Notice that the conditional probabilities themselves are very different: P(F \mid L) = 0.35 but P(L \mid F) = 0.7. About a third of the locals were fast, but most of the fast runners were local. Each one is paired with its own first factor.

Example 3

The cable car on a tree

On any day, the chance that it is windy at a mountain cable car is 0.25. On a windy day the cable car shuts with probability 0.6; on a calm day it shuts, for maintenance or other reasons, with probability 0.12. Find the probability that a day is windy and the cable car shuts, the probability that a day is calm and it shuts, and the probability that it shuts on a day chosen at random.

Solution

The first fan of the tree in Figure 2 is the weather: windy 0.25, calm 1 - 0.25 = 0.75. The second fans are the conditional probabilities. After windy: P(\text{shut} \mid \text{windy}) = 0.6 and P(\text{runs} \mid \text{windy}) = 0.4. After calm: P(\text{shut} \mid \text{calm}) = 0.12 and P(\text{runs} \mid \text{calm}) = 0.88. Each fan adds to 1, since given the weather the cable car either shuts or runs.

Figure 2Every second-level branch is a probability given the branch before it, and each path multiplies them.

Each path is the general multiplication rule:

\begin{aligned} P(\text{windy and shut}) &= 0.25 \times 0.6 = 0.15 \\ P(\text{calm and shut}) &= 0.75 \times 0.12 = 0.09 \end{aligned}

The cable car shuts along either of these two paths, and a day cannot be both windy and calm, so

P(\text{shut}) = 0.15 + 0.09 = 0.24

Check. The other two paths are 0.25 \times 0.4 = 0.10 and 0.75 \times 0.88 = 0.66, and 0.15 + 0.10 + 0.09 + 0.66 = 1. Adding up the paths that end in one result in this way is worked out in general in the law of total probability.

Example 4

When the condition changes nothing

One card is drawn from a shuffled standard deck of 52, which has 13 hearts and 4 kings. Let H be “a heart” and K be “a king”. Use the general multiplication rule to find P(H \text{ and } K), and decide whether H and K are independent. Then do the same with R, “a red card”, in place of K.

Solution

\displaystyle P(H) = \frac{13}{52} = \frac{1}{4}. Given that the card is a heart, it is one of 13 hearts, and exactly one of them is a king, so \displaystyle P(K \mid H) = \frac{1}{13}. The rule gives

P(H \text{ and } K) = \frac{1}{4} \times \frac{1}{13} = \frac{1}{52}

which is right, since exactly one card, the king of hearts, is both. Now compare: with nothing known, \displaystyle P(K) = \frac{4}{52} = \frac{1}{13}. So P(K \mid H) = P(K): knowing the suit does not change the chance of a king, and H and K are independent. The general rule has turned into P(H) \times P(K), the rule for independent events.

With R: every heart is red, so \displaystyle P(R \mid H) = \frac{13}{13} = 1, and

P(H \text{ and } R) = \frac{1}{4} \times 1 = \frac{1}{4}

which is right, since “a red heart” is just “a heart”. But \displaystyle P(R) = \frac{26}{52} = \frac{1}{2}, so P(R \mid H) \ne P(R), and H and R are dependent. Multiplying P(H) \times P(R) would give \displaystyle \frac{1}{8}, which is not the chance of anything here. The general rule gave the right answer in both cases; it did not need to be told in advance which case it was in.

Example 5

Three baton changes

A relay team of four runners makes three baton changes. The coach’s records give these chances: the first change is clean with probability 0.9; given that it was clean, the second is clean with probability 0.95; given that the first two were clean, the third is clean with probability 0.8. Find the probability that all three changes are clean, the probability that the first two are clean and the third is fumbled, and the probability that at least one change is fumbled.

Solution

Write A, B and C for “the first, second and third changes are clean”. The given numbers are P(A) = 0.9, P(B \mid A) = 0.95 and P(C \mid A \text{ and } B) = 0.8, which are exactly the three factors the rule for three events asks for:

\begin{aligned} P(A \text{ and } B \text{ and } C) &= 0.9 \times 0.95 \times 0.8 \\ &= 0.684 \end{aligned}

Given the first two were clean, the third is either clean or fumbled, so the chance it is fumbled is 1 - 0.8 = 0.2, and

\begin{aligned} &P(A \text{ and } B \text{ and not } C) \\ &\quad = 0.9 \times 0.95 \times 0.2 = 0.171 \end{aligned}

“At least one fumble” is everything except “all three clean”, so it is the complement:

1 - 0.684 = 0.316

Nothing here was drawn from a collection, and nothing was assumed independent: each factor is given everything before it, and that is all the rule needs.

Example 6

Two volunteers chosen at once

A group of 9 volunteers, 4 of them first-years and 5 of them not, chooses 2 people at random to speak at an event. The two are chosen at the same moment, not one after the other. Find the probability that both are first-years and the probability that the pair is one first-year and one other, and check both answers by counting pairs with combinations.

Solution

Choosing two at once gives the same pairs, each as likely as any other, as choosing them one at a time without repeats. So imagine the names read out one after the other, and let A be “the first name read is a first-year” and B be “the second name read is a first-year”. Then \displaystyle P(A) = \frac{4}{9} and \displaystyle P(B \mid A) = \frac{3}{8}, and

\begin{aligned} P(\text{both first-years}) &= \frac{4}{9} \times \frac{3}{8} \\ &= \frac{12}{72} = \frac{1}{6} \end{aligned}

One of each happens in two ways, first-year then other or other then first-year:

\begin{aligned} P(\text{one of each}) &= \frac{4}{9} \times \frac{5}{8} + \frac{5}{9} \times \frac{4}{8} \\ &= \frac{20}{72} + \frac{20}{72} = \frac{40}{72} = \frac{5}{9} \end{aligned}

Check by counting pairs. There are \binom{9}{2} = 36 pairs, all equally likely. Pairs of two first-years number \binom{4}{2} = 6, and \displaystyle \frac{6}{36} = \frac{1}{6}. Pairs of one of each number 4 \times 5 = 20, and \displaystyle \frac{20}{36} = \frac{5}{9}. The order of reading out was only a device; the rule works on it, and the answers do not depend on it.

Boundaries

The condition must have a chance. P(B \mid A) is defined only when P(A) > 0. If P(A) = 0 there is nothing to take a share of; but then “A and B” is impossible too, so P(A \text{ and } B) = 0 without the rule.

Each factor is given the events before it. P(A) pairs with P(B \mid A), and P(B) pairs with P(A \mid B). A product such as P(A) \times P(A \mid B) is not the chance of anything in general, as the race in Example 2 shows: 0.6 \times 0.7 = 0.42, but only 0.21 of the runners are local and fast. With three events, the third factor is given both earlier events, not only the one just before it.

A percentage has to say what it is a share of. “15\% cycle” may mean 15\% of everyone or 15\% of the train takers. Only the second is P(C \mid T). If it is 15\% of everyone, then 0.4 \times 0.15 is the chance of both only when the two events are independent.

The rule gives “and”, never “or”. The chance that at least one of A and B happens needs the general addition rule, which uses P(A \text{ and } B) as one of its parts.

Where this goes next

Turned round, the rule gives the conditional probability itself: when the chance of both and the chance of the first are known, \displaystyle P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)}. That is finding a conditional probability, and the idea of a probability given an event is studied in its own right in conditional probability.

Adding the paths of a tree that end in the same result, as in Example 3, grows into the law of total probability: split the possibilities into non-overlapping cases, multiply each case’s chance by the chance of the event given that case, and add. That in turn is the step that lets a conditional probability be turned round, from P(B \mid A) to P(A \mid B).

Exercises

Two events

P(A) = 0.8 and P(B \mid A) = 0.45. Find P(A \text{ and } B).
A drawer holds 14 socks, 6 of them black. Two socks are pulled out at random, one after the other, without putting the first back. Find the probability that both are black.
At a cinema, 65\% of customers buy a snack, and of those who buy a snack, 40\% also buy a drink. Find the probability that a customer chosen at random buys a snack and a drink.
A basket of 20 apples has 3 bruised ones. Two apples are taken at random without replacement. Find the probability that the first is bruised and the second is not, and the probability that the first is not bruised and the second is.
Let R be “it rains” and L be “the bus is late”. Write each statement in symbols: (a) on 30\% of rainy days the bus is late; (b) on 30\% of all days it rains and the bus is late; (c) of the days the bus is late, 30\% are rainy.

Either order

Of 250 people surveyed, 100 own a tablet, 160 own a laptop and 70 own both. One is chosen at random; B is “owns a tablet” and C is “owns a laptop”. Find P(B) \times P(C \mid B) and P(C) \times P(B \mid C), and check them against the count of people who own both.
P(A) = 0.5, P(B) = 0.6 and P(B \mid A) = 0.3. Find P(A \text{ and } B), and then use the rule in the other order to find P(A \mid B).
A factory’s parts come from two machines, and 20\% come from machine M. A part from machine M is faulty with probability 0.05. Find the probability that a part chosen at random came from machine M and is faulty, and the probability that it came from machine M and is not faulty.

Trees

The chance of a frost on a winter night is 0.3. Given a frost, a garden pipe bursts with probability 0.1; given no frost, with probability 0.01. Find the probabilities of the four paths of the tree and check that they add to 1.
A spinner lands on red with probability \displaystyle \frac{1}{3}. If it lands on red, a counter is drawn from bag X, which holds 2 gold counters among 5; otherwise from bag Y, which holds 1 gold counter among 4. Find the probability of red and gold, of not red and gold, and of gold.
On a tree, the first branch to A carries 0.6 and the path to “A and B” ends in 0.45. Find the probability on the branch from A to B, and the probability at the end of the path to “A and not B”.

Three or more events

Three packets are drawn without replacement from the jar of Example 1: 18 packets, 5 of them tomato. Find the probability that all three are tomato.
A walking group’s day out needs three things. The train runs with probability 0.95; given that, the footpath is open with probability 0.9; given both, the weather stays dry with probability 0.7. Find the probability that all three go right.
Three cards are dealt from a shuffled standard deck of 52. Find the probability that all three are hearts.

Independent or not

One card is drawn from a standard deck. S is “a spade” and F is “a picture card” (a jack, queen or king). Find P(F \mid S) and P(F), say whether S and F are independent, and find P(S \text{ and } F).
P(A) = 0.3 and P(B) = 0.5. Use the general multiplication rule to find P(B \mid A), and say whether A and B are independent, when (a) P(A \text{ and } B) = 0.15 and (b) P(A \text{ and } B) = 0.1.

Reasoning

Explain why P(A) \times P(B \mid A) = P(B) \times P(A \mid B) for any two events with positive probabilities, even though the factors on the two sides are usually different.
Use the general multiplication rule to explain why P(A \text{ and } B) can never be bigger than P(A).
A and B are mutually exclusive and P(A) > 0. What is P(B \mid A)? What does the general multiplication rule then give for P(A \text{ and } B), and why is that right?
A report says “40\% of commuters take the train and 15\% cycle”, and someone works out 0.4 \times 0.15 = 0.06 as the chance that a commuter does both. Say what the 15\% must mean for this to be right with no further assumption, and what has to be true if it means 15\% of all commuters.
Show that if P(A) > 0 and P(A \text{ and } B) = P(A) \times P(B), then P(B \mid A) = P(B).

Answers

0.36. By the rule, 0.8 \times 0.45 = 0.36.
\displaystyle \frac{15}{91}. \displaystyle \frac{6}{14} \times \frac{5}{13} = \frac{30}{182} = \frac{15}{91}: after a black sock, 5 of the 13 left are black.
0.26. P(\text{snack}) = 0.65 and P(\text{drink} \mid \text{snack}) = 0.4, so 0.65 \times 0.4 = 0.26.
Both \displaystyle \frac{51}{380}. Bruised then not: \displaystyle \frac{3}{20} \times \frac{17}{19} = \frac{51}{380}. Not then bruised: \displaystyle \frac{17}{20} \times \frac{3}{19} = \frac{51}{380}.
(a) P(L \mid R) = 0.3, a share of the rainy days. (b) P(R \text{ and } L) = 0.3, a share of all days. (c) P(R \mid L) = 0.3, a share of the late days.
Both 0.28. First, \displaystyle P(B) = \frac{100}{250} = 0.4 and \displaystyle P(C \mid B) = \frac{70}{100} = 0.7, so P(B) \times P(C \mid B) = 0.4 \times 0.7 = 0.28. Second, \displaystyle P(C) = \frac{160}{250} = 0.64 and \displaystyle P(B \mid C) = \frac{70}{160} = 0.4375, so the product is 0.64 \times 0.4375 = 0.28. The count gives \displaystyle \frac{70}{250} = 0.28.
P(A \text{ and } B) = 0.15 and P(A \mid B) = 0.25. First 0.5 \times 0.3 = 0.15. In the other order 0.6 \times P(A \mid B) = 0.15, so \displaystyle P(A \mid B) = \frac{0.15}{0.6} = 0.25.
0.01 and 0.19. 0.2 \times 0.05 = 0.01; given machine M, not faulty has probability 0.95, and 0.2 \times 0.95 = 0.19.
Frost and burst 0.03, frost and no burst 0.27, no frost and burst 0.007, no frost and no burst 0.693. These are 0.3 \times 0.1, 0.3 \times 0.9, 0.7 \times 0.01 and 0.7 \times 0.99, and 0.03 + 0.27 + 0.007 + 0.693 = 1.
Red and gold \displaystyle \frac{2}{15}, not red and gold \displaystyle \frac{1}{6}, gold \displaystyle \frac{3}{10}. \displaystyle \frac{1}{3} \times \frac{2}{5} = \frac{2}{15} and \displaystyle \frac{2}{3} \times \frac{1}{4} = \frac{1}{6}; the two paths to gold add to \displaystyle \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10}.
0.75 and 0.15. From 0.6 \times P(B \mid A) = 0.45, P(B \mid A) = 0.75; then P(\text{not } B \mid A) = 0.25 and 0.6 \times 0.25 = 0.15.
\displaystyle \frac{5}{408}, about 0.012. The product is \displaystyle \frac{5}{18} \times \frac{4}{17} \times \frac{3}{16}, which is \displaystyle \frac{60}{4896} = \frac{5}{408}.
0.5985. 0.95 \times 0.9 \times 0.7 = 0.5985, each factor given the ones before it.
\displaystyle \frac{11}{850}, about 0.013. The product is \displaystyle \frac{13}{52} \times \frac{12}{51} \times \frac{11}{50}, which is \displaystyle \frac{1716}{132600} = \frac{11}{850}.
\displaystyle P(F \mid S) = \frac{3}{13} and \displaystyle P(F) = \frac{12}{52} = \frac{3}{13}, so they are independent; \displaystyle P(S \text{ and } F) = \frac{1}{4} \times \frac{3}{13} = \frac{3}{52}, the three picture cards in spades.
(a) P(B \mid A) = 0.5, independent. From 0.3 \times P(B \mid A) = 0.15, and 0.5 = P(B). (b) \displaystyle P(B \mid A) = \frac{1}{3}, dependent. From 0.3 \times P(B \mid A) = 0.1, and \displaystyle \frac{1}{3} \ne 0.5.
Both sides equal P(A \text{ and } B), by the rule applied with A first and with B first, and “A and B” is the same event as “B and A”. The factors split the same probability in two different ways: the size of A times the share of B inside it, or the size of B times the share of A inside it.
P(B \mid A) is a probability, so it is at most 1, and P(A) \ge 0. So P(A) \times P(B \mid A) \le P(A) \times 1 = P(A). In words, the outcomes where both happen are some of the A outcomes, never more than all of them.
P(B \mid A) = 0, since when A has happened B cannot have. The rule gives P(A \text{ and } B) = P(A) \times 0 = 0, which is right, because mutually exclusive events cannot happen together.
The 15\% must be a share of the train takers, P(C \mid T) = 0.15; then the rule gives 0.06 directly. If it is 15\% of all commuters, it is P(C), and P(T) \times P(C) is the chance of both only if P(C \mid T) = P(C), that is if taking the train and cycling are independent.
The general rule gives P(A \text{ and } B) = P(A) \times P(B \mid A), and we are told P(A \text{ and } B) = P(A) \times P(B). So P(A) \times P(B \mid A) = P(A) \times P(B), and dividing by P(A), which is allowed because P(A) > 0, gives P(B \mid A) = P(B).