Dependent events and drawing without replacement
Two prizes are drawn from a hat of names, and nobody may win twice. Once the first name is out of the hat, what has happened to everybody else’s chances?
Where you are 10 in Compound events
Why this exists
The multiplication rule for independent events multiplies two chances because the second stage does not care how the first one went. A coin does not remember the last toss.
Plenty of everyday draws are not like that. Two raffle prizes are drawn from a hat of names, and a name that has won is kept out so that nobody wins twice. A coach picks a captain, then a vice-captain from the players who are left. Someone eats a chocolate from a box, then takes another. Each time the first draw takes something away, so the second is made from a smaller, different collection, and its chances are no longer those of the first.
This lesson finds the new chances by recounting what is left, carries them on a tree, compares drawing with and without putting back, and checks the tree against a plain count of outcomes.
Definition
Drawing with replacement means each item drawn is put back and mixed in before the next draw. Every draw is made from the same full collection.
Drawing without replacement means each item drawn is kept out. Each draw is made from what the earlier draws left behind, so the collection has one item fewer every time.
Two events are dependent when knowing how one turned out changes the probability of the other. They are independent when it does not, which is the case the multiplication rule handles.
Drawing without replacement makes the draws dependent. A box holds 10 chocolates, 4 caramel and 6 mint, all wrapped alike, and you take one without looking. The chance it is caramel is \displaystyle \frac{4}{10}. If it was caramel, the box now holds 9 chocolates and 3 of them are caramel, so the chance the next one is caramel is \displaystyle \frac{3}{9}. If it was mint, 4 caramels remain among 9, and the chance is \displaystyle \frac{4}{9}. The chance for the second draw depends on the result of the first, which is exactly what dependent means.
Had the first chocolate been put back and the box shaken, the second draw would see the same 10 chocolates with the same 4 caramels whatever came first, and the draws would be independent.
Intuition
For a draw from a well-mixed collection, a probability is a count: items of the kind you want over items there are. Without replacement the counts change, so recount before every draw. The total always drops by 1; the count of a kind drops by 1 only if that kind was the one just taken.
A tree diagram holds all of this recounting at once. The first fan of branches gives the chances for the first draw, from the full box. Each first-level branch then has its own second fan, worked out from what that particular first draw left behind. So the two second-level fans of a without-replacement tree carry different numbers, while on a with-replacement tree they are copies of the first fan. Figure 1 is the tree for taking two chocolates from the box above, with C for caramel and M for mint.
Read the top path: caramel first, with chance \displaystyle \frac{4}{10}; then, from a box with 3 caramels among 9, caramel again with chance \displaystyle \frac{3}{9}. The two numbers are multiplied because the second is a fraction of the cases where the first draw has already gone that way: in \displaystyle \frac{4}{10} of all tries the first chocolate is caramel, and in \displaystyle \frac{3}{9} of those tries the second is caramel too. A fraction of a fraction is a product.
Results
Recount what is left. When items are drawn without replacement from a well-mixed collection, and the earlier draws are known, the probability that the next item is of a given kind is
Multiply along a path, with the updated probabilities. For two draws,
and for three draws a third factor joins, recounted from what the first two draws left.
The fans add to one. The branches leaving any one point of a tree add to 1, since they cover every item still in. The probabilities at the ends of all the paths add to 1 as well.
An event of several paths. An event such as “one of each” is made of more than one path. Its probability is the sum of the path probabilities, since no two paths can happen together.
The counting check. When n items are drawn two at a time without replacement there are n(n-1) equally likely ordered outcomes, and a path’s probability is the number of ordered outcomes on that path divided by n(n-1).
Why it is true
Label the 10 chocolates of Figure 1 so they can be told apart: caramels 1 to 4 and mints 5 to 10. An outcome of two draws is an ordered pair: which chocolate came first and which came second. The first can be any of 10. For each of those, the second can be any of the 9 left, but not the same chocolate again. By the counting principle there are
ordered pairs. The box is well mixed and nobody looks, so no chocolate is favored on either draw and no pair is favored over another: the 90 ordered pairs are equally likely.
Now count the pairs on the top path, caramel then caramel. There are 4 choices of caramel for the first draw and then 3 caramels left for the second, so 4 \times 3 = 12 pairs. Its probability is \displaystyle \frac{12}{90}, and
The left side is the count of pairs on the path over the count of all pairs. The right side is the product along the path. They are the same fraction, split two ways. The same split works on every path: the numerators multiply to the number of ordered pairs on the path, and the denominators multiply to the number of all ordered pairs. With three draws the argument is the same with 10 \times 9 \times 8 ordered triples.
Worked examples
The lesson is taught in two parts. Examples 1 to 3 belong to the first: recounting what is left, building the tree, and multiplying along its paths. Examples 4 to 6 belong to the second: the same collection drawn with and without replacement, three draws, and checking a tree against a count.
Choosing a captain, then a vice-captain
A squad of 12 players includes 5 defenders. The coach draws one name from a hat to be captain, keeps it out, and draws another to be vice-captain. Find the probability that the captain is a defender. Then find the probability that the vice-captain is a defender, first if the captain was a defender and then if the captain was not. Compare the three.
SolutionThe first draw is from all 12 names, 5 of them defenders, so
If the captain was a defender, that name is out of the hat. There are 11 names left, and only 4 of them are defenders, so the vice-captain is a defender with probability \displaystyle \frac{4}{11}.
If the captain was not a defender, the name taken out was someone else’s. There are still 11 names, but all 5 defenders are among them, so the probability is \displaystyle \frac{5}{11}.
To compare, put all three over 132:
A defender as captain makes a defender vice-captain less likely, and a non-defender as captain makes it more likely. The second draw’s chance depends on the first, so these draws are dependent.
Two chocolates, on a tree
Dorcas takes two chocolates, one after the other and without looking, from the box of Figure 1: 10 chocolates, 4 caramel and 6 mint. Find the probability that both are caramel, that both are mint, and that she gets one of each.
SolutionThe first fan comes from the full box: caramel \displaystyle \frac{4}{10}, mint \displaystyle \frac{6}{10}. The second fans are recounted. After a caramel, 9 remain with 3 caramel and 6 mint. After a mint, 9 remain with 4 caramel and 5 mint. Those are the numbers in Figure 1. Multiply along each path:
“One of each” happens along two paths, caramel then mint and mint then caramel, and they cannot both happen, so their probabilities add:
Check. The four paths cover every result, so they should add to 1: 12 + 24 + 24 + 30 = 90, and \displaystyle \frac{90}{90} = 1.
Who reads aloud
Six students, Algernon, Beatrix-Mae, Clement, Dorcas, Emmerich and Felicity, are in a reading group. The teacher picks one at random to read the first passage, then picks one of the other five at random to read the second. Find the probability that Felicity reads one of the two passages.
SolutionA tree with two kinds is enough: “Felicity” and “not Felicity”. There are two paths on which she reads. She reads first, with probability \displaystyle \frac{1}{6}. Or someone else reads first, with probability \displaystyle \frac{5}{6}, and then she is one of the 5 left, with probability \displaystyle \frac{1}{5}:
Check. The only other path passes her over twice: \displaystyle \frac{5}{6} \times \frac{4}{5} = \frac{2}{3}, and \displaystyle \frac{1}{3} + \frac{2}{3} = 1. It also makes sense on its own: two of the six students read and nobody is favored, so each has a \displaystyle \frac{2}{6} chance of being one of them.
The same hat, drawn two ways
A hat holds the names of 8 musicians, 3 violinists and 5 cellists. Two prizes are drawn. In the first raffle each name drawn is put back, so one person could win both prizes; in the second it is kept out. For each raffle, find the probability that both prizes go to violinists and the probability that one goes to a violinist and one to a cellist. Figure 2 shows both trees.
Put back. The draws are independent, as in the multiplication rule, and every fan is the first fan again: violinist \displaystyle \frac{3}{8}, cellist \displaystyle \frac{5}{8}.
Kept out. The second fans are recounted from 7 names. After a violinist, 2 violinists and 5 cellists are left; after a cellist, 3 violinists and 4 cellists.
As decimals, both prizes to violinists is about 0.141 with the name put back and about 0.107 with it kept out, while one of each is about 0.469 put back and about 0.536 kept out.
Keeping the name out makes “two of the same kind” less likely and “one of each” more likely. The reason is in the recount: once a violinist has won, there is one violinist fewer to win again, but every cellist is still in. Each path’s numerators show it: 3 \times 2 in place of 3 \times 3, but 3 \times 5 either way.
Three bulbs from one box
A box of 10 light bulbs has 2 faulty ones. A tester takes 3 bulbs, one at a time, and keeps each out. Find the probability that none of the three is faulty, and that at least one is. Compare the second answer with what it would be if each bulb were put back before the next was taken. What is the probability that all three are faulty?
SolutionWrite G for a good bulb. None faulty is the single path G, G, G. Recount before each draw: 8 good of 10, then 7 good of 9, then 6 good of 8.
“At least one faulty” covers every other path, so it is the complement of none faulty:
Put back, every draw is \displaystyle \frac{8}{10} good, so none faulty is \displaystyle \left(\frac{8}{10}\right)^3 = \frac{64}{125} and at least one faulty is \displaystyle \frac{61}{125} = 0.488. Keeping the bulbs out makes catching a faulty one more likely: each good bulb taken leaves the faulty ones a bigger share of what remains.
All three faulty is impossible. After two faulty bulbs come out there are 0 faulty among the 8 left, so the third branch carries \displaystyle \frac{0}{8} = 0:
With replacement it would be \displaystyle \left(\frac{2}{10}\right)^3 = \frac{1}{125}, small but possible.
Checking a tree with a table
A small stack of 5 cards holds 2 hearts, H_1 and H_2, and 3 spades, S_1, S_2 and S_3. The stack is shuffled and two cards are dealt from the top. List every ordered outcome in a table, use it to find the probability of each of HH, HS, SH and SS, and check the answers against the tree.
SolutionRows give the first card and columns the second. A card cannot be dealt twice, so the five cells where row and column name the same card are marked “none”.
| H_1 | H_2 | S_1 | S_2 | S_3 | |
|---|---|---|---|---|---|
| H_1 | none | HH | HS | HS | HS |
| H_2 | HH | none | HS | HS | HS |
| S_1 | SH | SH | none | SS | SS |
| S_2 | SH | SH | SS | none | SS |
| S_3 | SH | SH | SS | SS | none |
That leaves 25 - 5 = 20 cells, which is 5 \times 4, and they are equally likely: the shuffle favors no card for the top, and none of the other four for the next. Counting cells gives HH in 2, HS in 6, SH in 6 and SS in 6, so
The tree, recounting after the first card, gives the same four numbers:
They agree, as the argument under “Why it is true” says they must: each numerator product counts the cells for that path, and 5 \times 4 counts all the cells. One of each, HS or SH, has probability \displaystyle \frac{12}{20} = \frac{3}{5}.
Boundaries
The collection must be well mixed, with every item still in equally likely. Recounting gives the probability only when no item is favored. If someone always picks the largest chocolate, or the names are not stirred, the fractions describe nothing.
The update uses what is known. A second-level branch assumes the first result is known. When it is not, the chance for the second draw comes from adding every path that ends in it, as in Exercise 19.
A kind can run out. A kind cannot be drawn more often than it appears. A branch whose kind is used up carries probability 0, as with the third faulty bulb in Example 5.
A large collection hides the change, but does not remove it. Drawing from 1{,}000 bulbs with 30 faulty, the second chance after a faulty first is \displaystyle \frac{29}{999}, very close to \displaystyle \frac{30}{1000} but not equal to it. The draws are still dependent, only slightly.
Not every repeat is a draw. Rolling a die twice, or spinning a spinner twice, takes nothing away, so nothing is recounted and the trials are independent.
Where this goes next
The second factor along a path, “the probability of B once A has happened”, is the idea of conditional probability, which gives it the symbol P(B \mid A), read “the probability of B given A”. With it, multiplying along a path becomes the general multiplication rule, which works for any two events, dependent or not, and the independent case is the one where the condition changes nothing.
Draws without replacement where the order does not matter, such as a hand of five cards or a committee of three, are handled by counting unordered groups, in probability with counting formulas. And the complement trick of Example 5, finding “at least one” from “none”, is worked out in full in at least one problems.
Exercises
Part 1: what is left after a draw
- A box holds 9 light bulbs, 2 of them faulty. Clement takes a bulb and sets it aside. Find the probability that the next bulb he takes is faulty, (a) if the first was faulty, and (b) if the first was good.
- A box of 11 chocolates has 4 dark and 7 milk. Two are taken without looking, and the first is eaten. Find the probability that the second is dark if the first was dark, and if the first was milk.
- For each pair of trials, say whether anything is kept out between them, and whether the chances on the second depend on the result of the first. (a) Drawing two raffle winners where nobody may win twice. (b) Rolling a die twice. (c) Choosing a captain and then a vice-captain from one squad. (d) Spinning a spinner twice. (e) Drawing a card, noting it, shuffling it back in, then drawing again.
Part 1: trees and paths
- A box of 6 chocolates has 2 coffee and 4 orange. Felicity takes two, one after the other. Find the probability that both are coffee, that both are orange, and that she gets one of each.
- A stack of 7 cards has 4 red and 3 black. It is shuffled and two cards are dealt. Find the probability of red then black, of black then red, and of one card of each color.
- A class of 8 students includes Emmerich and Dorcas. One student is picked at random to read aloud, then a different one for the next passage. Find the probability that Emmerich reads first and Dorcas second, and the probability that the two readers are Emmerich and Dorcas in either order.
- A box of 10 bulbs has 4 faulty ones. A tester takes bulbs one at a time, keeping each out, until he finds a good one. Find the probability that the first good bulb is the third one taken.
- A hat holds 8 names: 5 who live on Ashgrove Street and 3 on Birch Lane. Two names are drawn without replacement. The first fan of the tree reads Ashgrove \displaystyle \frac{5}{8}, Birch \displaystyle \frac{3}{8}. Give the four second-level branch probabilities: Ashgrove and Birch after an Ashgrove first draw, and Ashgrove and Birch after a Birch first draw.
Part 2: with and without replacement
- A box of 5 bulbs has 1 faulty. Two bulbs are tested. Find the probability that the faulty bulb is among those tested if the bulbs are kept out, and the probability that at least one test finds a faulty bulb if each bulb is put back before the next is taken.
- A teacher picks three different students, one at a time, from a group of 7 that includes Clement. Find the probability that Clement is not picked, and that he is.
- In the group of Exercise 10, the teacher now picks three times and allows repeats, so the same student could be picked more than once. Find the probability that Clement is never picked, and say whether he is more or less likely to be missed than in Exercise 10.
- A box of 9 chocolates has 3 mint and 6 toffee. Dorcas eats three, one after the other, without looking. Find the probability that all three are toffee, and that at least one is mint.
- A stack of 6 cards has 3 hearts and 3 spades. Three cards are dealt. Find the probability that all three are the same suit. Then find it if each card is shuffled back before the next is dealt.
- A box of 8 bulbs has 3 faulty. Three are taken and kept out. Find the probability that exactly two of them are faulty.
Part 2: checking a tree by counting
- A stack of 6 cards has 1 ace. Two cards are dealt. How many ordered outcomes are there, how many of them contain the ace, and what is the probability of getting the ace? Check the answer with a tree.
- A captain and then a vice-captain are drawn from a squad of 5: Algernon, Beatrix-Mae, Clement, Dorcas and Emmerich. How many ordered outcomes are there, in how many does Algernon hold one of the two posts, and what is the probability that he does?
- A stack of 5 cards has 2 hearts. Two cards are dealt. A table of ordered outcomes with replacement has 5 \times 5 cells; without replacement it has 5 \times 4. In each table, count the cells where both cards are hearts, and find the probability of two hearts.
Reasoning
- Explain why the branches leaving any one point of a tree for draws without replacement must add to 1, even though the second fans differ from each other.
- Algernon says: “Without replacement, the second chocolate from the box of Figure 1 is always less likely to be caramel than the first.” Explain when he is right, when he is wrong, and what the chance is when the first flavor is not known.
- In the table of Example 6, explain why the five cells marked “none” are left out, and why each of the other 20 cells is equally likely.
Answers
- (a) \displaystyle \frac{1}{8}: 8 bulbs remain and 1 of them is faulty. (b) \displaystyle \frac{2}{8} = \frac{1}{4}: 8 remain and both faulty ones are among them.
- \displaystyle \frac{3}{10} after a dark first, and \displaystyle \frac{4}{10} = \frac{2}{5} after a milk first. In both cases 10 are left; a dark first leaves 3 dark, a milk first leaves all 4.
- (a) Kept out; dependent, since a winner’s name is no longer in. (b) Nothing is removed; independent. (c) Kept out; dependent. (d) Nothing is removed; independent. (e) Put back; independent, since the second draw is from the full stack again.
- Both coffee \displaystyle \frac{1}{15}, both orange \displaystyle \frac{2}{5}, one of each \displaystyle \frac{8}{15}. Coffee then coffee is \displaystyle \frac{2}{6} \times \frac{1}{5} = \frac{2}{30}; orange then orange is \displaystyle \frac{4}{6} \times \frac{3}{5} = \frac{12}{30}; one of each is \displaystyle \frac{2}{6} \times \frac{4}{5} + \frac{4}{6} \times \frac{2}{5} = \frac{16}{30}. Check: 2 + 12 + 16 = 30.
- Red then black \displaystyle \frac{2}{7}, black then red \displaystyle \frac{2}{7}, one of each \displaystyle \frac{4}{7}. The paths are \displaystyle \frac{4}{7} \times \frac{3}{6} = \frac{12}{42} and \displaystyle \frac{3}{7} \times \frac{4}{6} = \frac{12}{42}, and one of each is their sum, \displaystyle \frac{24}{42}.
- Emmerich then Dorcas \displaystyle \frac{1}{56}; the pair in either order \displaystyle \frac{1}{28}. The first is \displaystyle \frac{1}{8} \times \frac{1}{7}; Dorcas then Emmerich is another \displaystyle \frac{1}{56}, and \displaystyle \frac{2}{56} = \frac{1}{28}.
- \displaystyle \frac{1}{10}. The path is faulty, faulty, good: \displaystyle \frac{4}{10} \times \frac{3}{9} \times \frac{6}{8} = \frac{72}{720}.
- After Ashgrove: Ashgrove \displaystyle \frac{4}{7}, Birch \displaystyle \frac{3}{7}. After Birch: Ashgrove \displaystyle \frac{5}{7}, Birch \displaystyle \frac{2}{7}. Each fan is recounted from the 7 names left, and each fan adds to 1.
- Kept out \displaystyle \frac{2}{5}, put back \displaystyle \frac{9}{25}. Kept out, both good is \displaystyle \frac{4}{5} \times \frac{3}{4} = \frac{3}{5}, so the faulty one is found with probability \displaystyle 1 - \frac{3}{5}. Put back, both good is \displaystyle \left(\frac{4}{5}\right)^2 = \frac{16}{25}, and \displaystyle 1 - \frac{16}{25} = \frac{9}{25}.
- Not picked \displaystyle \frac{4}{7}, picked \displaystyle \frac{3}{7}. Missed three times is \displaystyle \frac{6}{7} \times \frac{5}{6} \times \frac{4}{5} = \frac{120}{210}; the rest is \displaystyle 1 - \frac{4}{7}.
- Never picked \displaystyle \frac{216}{343}, so he is more likely to be missed. Each pick misses him with probability \displaystyle \frac{6}{7}, so \displaystyle \left(\frac{6}{7}\right)^3 = \frac{216}{343}, while \displaystyle \frac{4}{7} = \frac{196}{343}. With repeats allowed, a pick can land on someone already chosen, which uses up a turn without reaching Clement.
- All toffee \displaystyle \frac{5}{21}, at least one mint \displaystyle \frac{16}{21}. All toffee is \displaystyle \frac{6}{9} \times \frac{5}{8} \times \frac{4}{7} = \frac{120}{504}, and at least one mint is its complement.
- Without replacement \displaystyle \frac{1}{10}, with replacement \displaystyle \frac{1}{4}. Three hearts is \displaystyle \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} = \frac{6}{120}, and three spades the same, so the total is \displaystyle \frac{12}{120}. Put back, each is \displaystyle \left(\frac{1}{2}\right)^3 = \frac{1}{8}, and \displaystyle \frac{2}{8} = \frac{1}{4}.
- \displaystyle \frac{15}{56}. Three paths give exactly two faulty: FFG, FGF and GFF. Each has numerators 3, 2 and 5 in some order over 8 \times 7 \times 6, so each is \displaystyle \frac{30}{336}, and \displaystyle \frac{90}{336} = \frac{15}{56}.
- 30 ordered outcomes, 10 with the ace, probability \displaystyle \frac{1}{3}. There are 6 \times 5 = 30; the ace is first in 5 and second in 5. Tree: ace first \displaystyle \frac{1}{6}, or not first and then the ace, \displaystyle \frac{5}{6} \times \frac{1}{5} = \frac{1}{6}; together \displaystyle \frac{2}{6} = \frac{1}{3}.
- 20 ordered outcomes, 8 with Algernon, probability \displaystyle \frac{2}{5}. There are 5 \times 4 = 20; Algernon is captain in 4 and vice-captain in 4, and \displaystyle \frac{8}{20} = \frac{2}{5}.
- With replacement 4 cells, probability \displaystyle \frac{4}{25}; without, 2 cells, probability \displaystyle \frac{2}{20} = \frac{1}{10}. With replacement either heart can come first and either second, 2 \times 2; without, the second must be the other heart, 2 \times 1.
- The branches leaving one point list every kind of item still in, and their numerators are the counts of those kinds. The counts of all kinds add to the number of items still in, which is the shared denominator, so the fractions add to 1. Each fan comes from a different collection, but each is a complete list for its own collection.
- He is right when the first chocolate is known to be caramel: \displaystyle \frac{3}{9} is less than \displaystyle \frac{4}{10}. He is wrong when it is known to be mint: \displaystyle \frac{4}{9} is more than \displaystyle \frac{4}{10}. When the first flavor is not known, add the two paths of Figure 1 that end in caramel: \displaystyle \frac{12}{90} + \frac{24}{90} = \frac{36}{90} = \frac{4}{10}, exactly the chance for the first draw. The draws are still dependent, because knowing the first flavor changes the answer.
- A cell marked “none” would mean the same card was dealt twice, and once a card is off the stack it cannot be dealt again. For the rest: a shuffled stack puts each of the 5 cards on top with the same chance, and then each of the 4 left next with the same chance, so every allowed ordered pair has the same chance, \displaystyle \frac{1}{20}.