Math 7 · 7.SP.C.8 · S.CP.A.2

Independent events and the multiplication rule

A coin is tossed and a number cube is rolled. Knowing how the coin landed tells you nothing about the cube, so how likely is it that both come out the way you want?

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Why this exists

Many chance experiments happen in stages: toss a coin and then roll a number cube, spin a spinner twice, draw a card and then draw again. A table of every outcome, as in Listing a sample space: lists and tables, answers any question about such an experiment, but the tables grow fast. Two spins of a ten-section spinner give 100 outcomes; three rolls of a number cube give 216.

Often the stages have nothing to do with each other: the coin does not know what the cube will do. Then the chance that both stages go a certain way is the chance for the first times the chance for the second. This lesson says when that short cut is allowed, shows why it agrees with counting, and stretches it to three or more stages.

Definition

Two events are independent when knowing whether one of them happened does not change the probability of the other. If they are not independent, they are dependent.

Independence usually comes from stages that cannot affect each other. A coin toss and a cube roll are independent. So are two spins of the same spinner, because the spinner has the same sections both times.

Sampling with replacement means drawing one item from a collection, noting it, putting it back, mixing, and drawing again. The second draw is made from exactly the same collection as the first, so its chances are the same whatever came first: draws made with replacement are independent. If the item is kept out instead, the collection has changed and the draws are dependent; that case belongs to Dependent events and drawing without replacement.

The event “A and B” is the event that both A and B happen, and its probability is written P(A \text{ and } B). Here the two events usually come from two stages: “the coin shows heads and the cube shows a 6”.

Intuition

Take the spinner in Figure 1. Its five sections are equal, so each is equally likely, and red, being one section out of five, has probability \displaystyle \frac{1}{5}. Blue and yellow each cover two sections, so each has probability \displaystyle \frac{2}{5}.

Figure 1Five equal sections, so red has probability \displaystyle \frac{1}{5}, blue \displaystyle \frac{2}{5} and yellow \displaystyle \frac{2}{5}.

Spin it twice. Pair each of the five sections for the first spin with each of the five for the second and you get the 5 \times 5 = 25 squares of Figure 2, all equally likely.

Figure 2Spinning twice. Each row is one section for the first spin, each column one section for the second. The shaded squares are “blue, then yellow”.

Look along any row. Whatever the first spin did, the row holds one red square, two blue and two yellow. The second spin’s chances, \displaystyle \frac{1}{5}, \displaystyle \frac{2}{5} and \displaystyle \frac{2}{5}, are the same in every row. That is what independence looks like in a table: learning which row you are in tells you nothing new about the column.

Now count “blue, then yellow”. Blue takes 2 of the 5 rows. Inside each of those rows, yellow takes 2 of the 5 squares. So the event is \displaystyle \frac{2}{5} of \displaystyle \frac{2}{5} of the grid, and “of” with fractions means multiply (Multiplying fractions):

\frac{2}{5} \times \frac{2}{5} = \frac{4}{25}

The shaded squares agree: 4 of the 25.

Results

The multiplication rule for independent events. If A and B are independent, then

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

Three or more independent events. If A, B and C are independent (none of them changes the chances of the others), then

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

and the same holds for four events or more: multiply all of their probabilities.

The same chance, repeated. If one event has probability p on each try and the tries are independent, then the probability that it happens on all of n tries is

p \times p \times \cdots \times p = p^n

Why it is true

Suppose the first stage has m equally likely results and the second has n. Draw a table with a row for each result of the first stage and a column for each result of the second. It has m \times n squares, one for each way the two stages can go.

Here is where independence is used. Because the first stage does not change the second, every column is just as likely in every row, so all m \times n squares are equally likely. The probability of an event is then the number of its squares over m \times n, as in Probability of an event.

Let A be an event about the first stage that covers a of the rows, and B an event about the second stage that covers b of the columns. Then

P(A) = \frac{a}{m} \qquad P(B) = \frac{b}{n}

The squares where both happen are the ones in an A row and a B column. There are a such rows with b such squares in each, so a \times b squares, and

\begin{aligned} P(A \text{ and } B) &= \frac{a \times b}{m \times n} \\ &= \frac{a}{m} \times \frac{b}{n} \\ &= P(A) \times P(B) \end{aligned}

The middle step is the rule for multiplying fractions read backwards: A fills a fraction \displaystyle \frac{a}{m} of the rows, and inside each of those rows B fills a fraction \displaystyle \frac{b}{n}, so both together fill a fraction of a fraction. A decimal chance fits the same picture: \displaystyle 0.8 = \frac{4}{5} is like drawing one of five equally likely slips, four marked “yes”.

For three independent stages, stack the tables into a box of m \times n \times k equally likely cells. The cells where all three events happen number a \times b \times c, and the same step gives \displaystyle \frac{a}{m} \times \frac{b}{n} \times \frac{c}{k}.

Worked examples

Example 1

A coin and a number cube

A coin is tossed and a number cube is rolled. The table lists all 12 equally likely outcomes.

cube	heads	tails
1	H1	T1
2	H2	T2
3	H3	T3
4	H4	T4
5	H5	T5
6	H6	T6

(a) Use the table to decide whether the events “heads” and “the cube shows more than 4” are independent.

(b) Find P(\text{heads and more than } 4) by counting, then by the multiplication rule.

Solution

(a) “More than 4” means a 5 or a 6. In the whole table that is 4 of the 12 outcomes, a probability of \displaystyle \frac{4}{12} = \frac{1}{3}.

Now suppose you are told the coin showed heads. Only the heads column is left: 6 outcomes, of which H5 and H6 are more than 4. The probability is \displaystyle \frac{2}{6} = \frac{1}{3}. The tails column gives \displaystyle \frac{2}{6} = \frac{1}{3} as well. Learning the coin leaves the chance at \displaystyle \frac{1}{3}, so the two events are independent.

(b) Counting: the outcomes with heads and more than 4 are H5 and H6, so the probability is \displaystyle \frac{2}{12} = \frac{1}{6}.

Multiplying: \displaystyle P(\text{heads}) = \frac{1}{2} and \displaystyle P(\text{more than } 4) = \frac{1}{3}, so

P(\text{heads and more than } 4) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}

The two ways agree, as the argument in the previous section says they must.

Example 2

Spinning twice

The spinner in Figure 1 is spun twice. Find the probability of

(a) red on both spins;

(b) blue on the first spin and yellow on the second;

Solution

The two spins are independent: the spinner has the same five sections each time, and the first spin cannot move them.

(a) \displaystyle P(\text{red}) = \frac{1}{5} on each spin, so

P(\text{red, then red}) = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}

In Figure 2 this is the single square in the top left corner.

(b) \displaystyle P(\text{blue}) = \frac{2}{5} and \displaystyle P(\text{yellow}) = \frac{2}{5}, so

P(\text{blue, then yellow}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}

These are the 4 shaded squares.

(c) “One blue and one yellow” happens in two separate ways: blue then yellow, or yellow then blue. Each has probability \displaystyle \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}. No pair of spins is both at once, since the first spin cannot be blue and yellow together, so the two chances add:

\frac{4}{25} + \frac{4}{25} = \frac{8}{25}

In Figure 2 these are the 4 shaded squares and the 4 squares mirrored across the diagonal, 8 in all. Part (b) names an order and part (c) does not, which is why (c) is twice as likely.

Example 3

Drawing with replacement

A pile of ten cards numbered 1 to 10 is shuffled. Una draws a card, notes its number, puts it back, shuffles, and draws again. Find the probability that

(a) both cards are even;

(b) the first card is the 7 and the second card is 3 or less.

Solution

Because the first card goes back and the pile is shuffled, the second draw is made from the same ten cards as the first. The two draws are independent.

(a) The even cards are 2, 4, 6, 8 and 10, five of the ten, so each draw is even with probability \displaystyle \frac{5}{10} = \frac{1}{2}. Then

P(\text{even, then even}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

(b) The first card is the 7 with probability \displaystyle \frac{1}{10}. The cards 3 or less are 1, 2 and 3, so the second draw gives one of them with probability \displaystyle \frac{3}{10}. Then

P(7\text{, then } 3 \text{ or less}) = \frac{1}{10} \times \frac{3}{10} = \frac{3}{100}

A table would have 10 \times 10 = 100 squares, and this event is the 3 squares in the row for 7 and the columns for 1, 2 and 3.

Had Una kept the first card out, the second draw would come from nine cards: after an even first card only 4 of the 9 left are even, after an odd one 5 of the 9. The draws would be dependent.

Example 4

Two arrows

Xander hits the target with 80\% of his arrows. He shoots two, and we assume the shots are independent: hitting or missing with the first does not change his chance with the second. Find the probability that he

(a) hits with both;

(b) hits with the first and misses with the second;

(d) misses with both.

Solution

On each shot P(\text{hit}) = 0.8 and, by the complement rule from Complements and probabilities that add to one, P(\text{miss}) = 1 - 0.8 = 0.2.

(a) P(\text{hit, hit}) = 0.8 \times 0.8 = 0.64.

(b) P(\text{hit, miss}) = 0.8 \times 0.2 = 0.16.

0.8 \times 0.2 + 0.2 \times 0.8 = 0.16 + 0.16 = 0.32

(d) P(\text{miss, miss}) = 0.2 \times 0.2 = 0.04.

Check. Two shots must give two hits, one hit or no hits, and nothing else, so the three chances should add to 1: 0.64 + 0.32 + 0.04 = 1. They do.

Example 5

Three stages at once

(a) A coin is tossed, a number cube is rolled and the spinner in Figure 1 is spun. Find the probability of heads, a 6 and red.

(b) The spinner is spun three times. Find the probability that it never lands on red, and the probability that it lands on red at least once.

Solution

(a) The three stages cannot affect one another, so the rule for three independent events applies:

\begin{aligned} &P(\text{heads and } 6 \text{ and red}) \\ &\quad = \frac{1}{2} \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{60} \end{aligned}

Check by counting. The three stages have 2, 6 and 5 results, so there are 2 \times 6 \times 5 = 60 equally likely outcomes (see The fundamental counting principle), and exactly one of them is heads, 6, red.

(b) On each spin \displaystyle P(\text{not red}) = 1 - \frac{1}{5} = \frac{4}{5}. Three independent spins give

\begin{aligned} P(\text{no red}) &= \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \\ &= \left(\frac{4}{5}\right)^3 = \frac{64}{125} \end{aligned}

“At least one red” is the complement of “no red”: every set of three spins has either no red or some red, never both. So

P(\text{at least one red}) = 1 - \frac{64}{125} = \frac{61}{125}

Counting “at least one red” directly means adding one, two and three reds in every order; the complement is quicker. At least one problems take this further.

Boundaries

The events must be independent. The rule P(A \text{ and } B) = P(A) \times P(B) is not true for dependent events. Drawing without putting the first item back is the standard dependent case, and its rule multiplies by an updated chance instead: see Dependent events and drawing without replacement and, later, The general multiplication rule.

Two events on the same trial are not automatically independent. On one roll of a number cube, let A be “even” and B be “more than 3”. Then \displaystyle P(A) = \frac{1}{2} and \displaystyle P(B) = \frac{1}{2}, but both happen only on 4 and 6, so \displaystyle P(A \text{ and } B) = \frac{2}{6} = \frac{1}{3}, not \displaystyle \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. Knowing the roll was more than 3 raises the chance of even from \displaystyle \frac{1}{2} to \displaystyle \frac{2}{3}. Independence of events on the same trial has to be checked from the outcomes; it is never assumed.

In the real world, independence is an assumption. Coins, cubes and spinners are independent from one use to the next because nothing carries over. Arrows, seeds, machines and weather are a modeling choice: the rule gives the right answer only as far as the stages really do not influence each other. If one cause, such as a cold snap or a worn part, can push two events the same way, multiplying their chances understates how often both happen.

Where this goes next

Multiplying chances along a sequence of stages is the idea behind tree diagrams, where each branch carries a probability and a whole path is found by multiplying along it. When the stages do affect each other, the second chance is updated before multiplying, which is Dependent events and drawing without replacement; the general version, which needs Conditional probability, is the The general multiplication rule.

Repeating the same independent try many times leads to the The binomial distribution, which counts successes in a fixed number of independent tries; checking that the tries really are independent is one of the conditions in The binomial setting. And independence is why a coin has no memory: after a long run of heads the next toss is still \displaystyle \frac{1}{2}, the point made in The law of large numbers and the gambler's fallacy.

Exercises

Independent or not

Say whether the two events are independent, and give a reason. (a) A coin is tossed and a number cube is rolled. A: heads. B: the cube shows 3. (b) A card is drawn from ten cards numbered 1 to 10, put back, the pile shuffled, and a second card drawn. A: the first card is the 4. B: the second card is the 4. (c) The same as (b), but the first card is kept out. A: the first card is the 4. B: the second card is the 4. (d) A: it rains in a town this afternoon. B: the pavements there are wet this evening.
Use the coin and number cube table in Example 1. Find the probability that the cube shows an even number, first over the whole table and then only in the tails column. Are “tails” and “even” independent? Find P(\text{tails and even}) by counting.

Two independent events

The spinner in Figure 1 is spun twice. Find the probability of yellow on both spins.
The spinner in Figure 1 is spun twice. Find the probability of red first and blue second.
The spinner in Figure 1 is spun twice. Find the probability that neither spin is red.
A bean seed sprouts with probability 0.85. Two seeds are planted in separate pots, and they sprout or fail independently. Find the probability that (a) both sprout, (b) neither sprouts, (c) exactly one sprouts.
A printer misfeeds a page with probability 0.03 and, separately, smudges a page with probability 0.04. Assume the two are independent. Find the probability that one page is (a) misfed and smudged, (b) neither, (c) at least one of the two.

Three or more independent events

A coin is tossed four times. Find the probability of (a) four heads, (b) heads, tails, heads, tails in exactly that order.
The spinner in Figure 1 is spun three times. Find the probability of (a) red every time, (b) blue every time.
A coin is tossed, a number cube is rolled and the spinner in Figure 1 is spun. Find the probability of heads, an odd number and blue.
Three cards are drawn, with replacement, from ten cards numbered 1 to 10. Find the probability that all three are greater than 7.

Working backwards

Events A and B are independent, with P(A) = 0.6 and P(A \text{ and } B) = 0.24. Find P(B).
A spinner with 14 equal sections, some of them red, is spun twice. The probability of red on both spins is \displaystyle \frac{1}{49}. What is the probability of red on one spin, and how many sections are red?
A coin is tossed again and again. What is the smallest number of tosses for which the probability that every toss is heads is less than \displaystyle \frac{1}{100}?
Tancred tosses a coin and rolls a number cube. The probability of heads and a number less than k is \displaystyle \frac{1}{4}. Find the whole number k.

Explaining

Viggo tosses a coin five times and gets heads every time. He says tails is now more likely on the sixth toss. Explain why he is wrong, and compare the chances of HHHHHT and HHHHHH.
The spinner in Figure 1 is spun and a coin is tossed. Draw or describe a table of equally likely outcomes and use it to explain why \displaystyle P(\text{red and heads}) = \frac{1}{5} \times \frac{1}{2}.
Ottoline says two events on the same roll of a number cube can never be independent. Test her claim with A: “even” and B: “5 or more”.
Xander’s coach says that after a miss, Xander rushes his next arrow. Explain why that makes 0.8 \times 0.8 an unsafe answer for the chance that he hits with two arrows in a row, even though each of his arrows on its own hits 80\% of the time.

Answers

(a) Independent: the coin cannot change how the cube lands. (b) Independent: the card goes back and the pile is shuffled, so the second draw is from the same ten cards whatever the first was. (c) Dependent: if the first card is the 4, the second cannot be, a chance of 0; otherwise the chance is \displaystyle \frac{1}{9}. (d) Dependent: rain this afternoon makes wet pavements this evening much more likely.
Over the whole table, \displaystyle \frac{6}{12} = \frac{1}{2}. In the tails column, T2, T4, T6 give \displaystyle \frac{3}{6} = \frac{1}{2}. The same, so independent. \displaystyle P(\text{tails and even}) = \frac{3}{12} = \frac{1}{4}, which is \displaystyle \frac{1}{2} \times \frac{1}{2}.
\displaystyle \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}.
\displaystyle \frac{1}{5} \times \frac{2}{5} = \frac{2}{25}.
Not red has probability \displaystyle \frac{4}{5} each spin: \displaystyle \frac{4}{5} \times \frac{4}{5} = \frac{16}{25}.
(a) 0.85 \times 0.85 = 0.7225. (b) A seed fails with probability 1 - 0.85 = 0.15, so 0.15 \times 0.15 = 0.0225. (c) 0.85 \times 0.15 + 0.15 \times 0.85 = 0.255. Check: 0.7225 + 0.0225 + 0.255 = 1.
(a) 0.03 \times 0.04 = 0.0012. (b) 0.97 \times 0.96 = 0.9312. (c) The complement of “neither”: 1 - 0.9312 = 0.0688.
(a) \displaystyle \left(\frac{1}{2}\right)^4 = \frac{1}{16}. (b) Also \displaystyle \frac{1}{16}: each toss must show one named face, chance \displaystyle \frac{1}{2} each time.
(a) \displaystyle \left(\frac{1}{5}\right)^3 = \frac{1}{125}. (b) \displaystyle \left(\frac{2}{5}\right)^3 = \frac{8}{125}.
\displaystyle \frac{1}{2} \times \frac{3}{6} \times \frac{2}{5} = \frac{1}{10}.
The cards greater than 7 are 8, 9, 10, so \displaystyle \frac{3}{10} each draw: \displaystyle \left(\frac{3}{10}\right)^3 = \frac{27}{1000}.
0.6 \times P(B) = 0.24, so P(B) = 0.24 \div 0.6 = 0.4.
\displaystyle \frac{1}{7}, since \displaystyle \frac{1}{7} \times \frac{1}{7} = \frac{1}{49}. Then \displaystyle \frac{1}{7} of 14 sections is 2 red sections.
7 tosses. Six give \displaystyle \left(\frac{1}{2}\right)^6 = \frac{1}{64}, more than \displaystyle \frac{1}{100}; seven give \displaystyle \frac{1}{128}, less.
k = 4. \displaystyle \frac{1}{2} \times P(\text{less than } k) = \frac{1}{4} needs \displaystyle P(\text{less than } k) = \frac{1}{2}, three numbers out of six, which are 1, 2, 3.
The tosses are independent, so the sixth is tails with probability \displaystyle \frac{1}{2} whatever came before; the coin has no memory. HHHHHT and HHHHHH each have probability \displaystyle \left(\frac{1}{2}\right)^6 = \frac{1}{64}, equally likely.
Five rows for the five equal sections and two columns for heads and tails give 10 equally likely squares. Red is one row, and heads one square in it, so \displaystyle P = \frac{1}{10}, which is \displaystyle \frac{1}{5} of the rows times \displaystyle \frac{1}{2} of each row.
Her claim is false. \displaystyle P(A) = \frac{1}{2}, \displaystyle P(B) = \frac{2}{6} = \frac{1}{3}, and both happen only on 6, so \displaystyle P(A \text{ and } B) = \frac{1}{6} = \frac{1}{2} \times \frac{1}{3}: these two are independent. Half of \{5, 6\} is even, the same as half of the whole cube.
The rule needs his chance with the second arrow to be 0.8 whatever the first did. If a miss changes the next shot, the shots are dependent, and the chance of two hits has to use his rate on an arrow that follows a hit, which need not be 0.8.