Algebra 2

At least one problems

Six fireworks are set out for a show. What is the chance that at least one of them goes off, when counting every way that could happen would take all afternoon?

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When you reach for it

Ffion sets out 6 fireworks. Each one goes off with probability 0.85, and the fireworks do not affect one another. What is the chance that at least one of them goes off?

“At least one” is a crowded event. It holds every way of getting one firework to go off, two, three, four, five or six, and each of those can happen in many orders: exactly two going off alone can be any 2 of the 6 fireworks. Adding all of that up is a long job with many chances to slip.

Its opposite is short. The only way to fail to get at least one is for none of the fireworks to go off: firework one fails, and firework two fails, and so on to firework six. That is a single run of failures, and a run of independent events is one product. So the move is:

P(\text{at least one}) = 1 - P(\text{none})

Reach for it whenever a question says “at least one”, “one or more”, “any of them”, or “not all miss”: an alarm that must go off at least once, a message sent several ways that must arrive at least once, a test that must catch at least one fault. Reach for it again, the other way round, when you need to know how many tries it takes to make “at least one” likely enough, because the same formula with n tries puts the unknown in an exponent.

The procedure

Why “none” is the complement

Every outcome of the experiment has some number of successes: 0, or 1, or 2, and so on. An outcome with 0 successes is in “none”; an outcome with any other number is in “at least one”. No outcome is in both and none is left out, so the two events are complements, and complements have probabilities that add to 1:

P(\text{at least one}) + P(\text{none}) = 1

That is the whole reason the subtraction is allowed. It is worth saying exactly what the complement is each time, because it depends on the wording. The complement of “at least one goes off” is “no firework goes off”. The complement of “at least one fails to go off” is “every firework goes off”. Turning a question into its complement means negating the whole statement, not swapping one word.

Why “none” is a product

“None succeed” means the first try fails and the second fails and every other one fails. When the tries are independent, the multiplication rule says the chance of all of them together is the product of their separate chances. Each separate chance is a failure chance, which is itself a complement: a try that succeeds with probability p fails with probability 1 - p. So for n independent tries that each succeed with probability p,

\begin{aligned} P(\text{none}) &= (1 - p)^n\\ P(\text{at least one}) &= 1 - (1 - p)^n \end{aligned}

When the tries have different chances p_1, p_2, \ldots, p_n, the same reasoning multiplies the different failure chances:

\begin{aligned} P(\text{at least one}) &= 1 - (1 - p_1)(1 - p_2)\\ &\qquad \cdots(1 - p_n) \end{aligned}

When the tries are not independent, as when items are drawn and kept out, “none” is still one path through a tree, and its chance is still a product, but each factor is the updated chance of failing given that every earlier try failed, as in Dependent events and drawing without replacement.

Why not just add the chances

Adding p_1 + p_2 + \cdots + p_n would count an outcome where two tries succeed twice, once for each. The events “try 1 succeeds” and “try 2 succeeds” overlap, and the general addition rule has to remove every overlap. With six fireworks at 0.85 the sum is 5.1, which cannot be a probability at all. The product of failures has no overlaps to correct, because “none” is a single path.

How many tries for a target

Suppose each try succeeds with probability p and you want the chance of at least one success to be at least some target T. Then you need

1 - (1 - p)^n \ge T

which is the same as

(1 - p)^n \le 1 - T

The unknown n is an exponent, so take a logarithm of both sides. A logarithm with base 10 is an increasing function, so it keeps the direction of the inequality, and the power law brings n down:

n \log(1 - p) \le \log(1 - T)

Now divide by \log(1 - p). Because 1 - p is between 0 and 1, its logarithm is negative, and dividing an inequality by a negative number reverses it:

n \ge \frac{\log(1 - T)}{\log(1 - p)}

Since n counts tries, it must be a whole number, so take the next whole number up and check it and the one before it directly.

Worked examples

Example 1

Six fireworks

Ffion sets out 6 fireworks, each going off with probability 0.85, independently of the others. Find (a) the probability that none goes off, (b) the probability that at least one goes off, and (c) the probability that at least one fails to go off.

Solution

(a) Each firework fails with probability 1 - 0.85 = 0.15, by the complement rule. “None goes off” is all six failing, and the fireworks are independent, so the six failure chances multiply:

P(\text{none}) = 0.15^6 = 0.000011390625

(b) “At least one goes off” is the complement of “none goes off”:

\begin{aligned} P(\text{at least one}) &= 1 - 0.15^6\\ &= 0.999988609375 \approx 0.99999 \end{aligned}

Six fireworks at 0.85 each make a silent show almost impossible.

(c) This is a different question. The complement of “at least one fails” is “every firework goes off”, which is six successes in a row:

\begin{aligned} P(\text{all go off}) &= 0.85^6 = 0.377149515625\\ P(\text{at least one fails}) &= 1 - 0.85^6 \approx 0.623 \end{aligned}

So a perfect show of six is less likely than not, even though a silent one is next to impossible. Parts (b) and (c) both say “at least one”, and the difference lies entirely in what that “one” is.

Example 2

Three backup generators

A hospital has three backup generators that start independently with probabilities 0.9, 0.8 and 0.75. When the mains power fails, what is the probability that at least one generator starts?

Solution

The failure chances are the complements of the starting chances: 0.1, 0.2 and 0.25. “No generator starts” needs all three failures together, and the generators are independent, so

P(\text{none starts}) = 0.1 \times 0.2 \times 0.25 = 0.005

and therefore

P(\text{at least one starts}) = 1 - 0.005 = 0.995

Adding the starting chances gives 0.9 + 0.8 + 0.75 = 2.45, more than 1, which shows at once why adding is not the move: the events overlap. Notice also what the backups buy. The best single generator leaves a 0.1 chance of no power; three together leave 0.005, twenty times smaller.

Example 3

At least one six

(a) A fair number cube is rolled 4 times. Find the probability of at least one six.

(b) A pair of fair number cubes is rolled 24 times. Find the probability of at least one double six, and compare the two answers.

Solution

(a) A roll misses six with probability \displaystyle \frac{5}{6}, and the rolls are independent, so

\begin{aligned} P(\text{no six}) &= \left(\frac{5}{6}\right)^4 = \frac{625}{1296}\\ P(\text{at least one six}) &= 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518 \end{aligned}

(b) The two cubes give 6 \times 6 = 36 equally likely pairs and only one of them is a double six, so each throw of the pair gives a double six with probability \displaystyle \frac{1}{36} and misses with probability \displaystyle \frac{35}{36}. Twenty-four independent throws give

\begin{aligned} P(\text{at least one double six}) &= 1 - \left(\frac{35}{36}\right)^{24}\\ &\approx 0.491 \end{aligned}

In both games the number of tries times the chance per try is the same: \displaystyle 4 \times \frac{1}{6} = \frac{2}{3} and \displaystyle 24 \times \frac{1}{36} = \frac{2}{3}. Yet the first game gives a little better than an even chance and the second a little worse. So the chance of at least one success is not “tries times chance”; that product only counts how many successes come on average. The complement gives the true answer.

Example 4

How many calls

A caller rings a busy helpline. Each call gets through with probability 0.1, independently of the others. How many calls must the caller be prepared to make so that the probability of getting through at least once is at least 0.9?

Solution

With n calls, each failing with probability 0.9, the chance of getting through at least once is 1 - 0.9^n. The caller needs

1 - 0.9^n \ge 0.9

which is the same as

0.9^n \le 0.1

by subtracting 1 from both sides, which gives -0.9^n \ge -0.1, and then multiplying both sides by -1, which reverses the inequality. The unknown is an exponent, so take the logarithm of both sides. The logarithm is increasing, so the inequality keeps its direction, and the power law brings n down:

n \log 0.9 \le \log 0.1

Here \log 0.9 \approx -0.0458 is negative, because 0.9 < 1. Dividing both sides by it reverses the inequality:

n \ge \frac{\log 0.1}{\log 0.9} \approx 21.85

The number of calls is a whole number, so n = 22. Check both sides of the boundary: 0.9^{21} \approx 0.109, which is too big, and 0.9^{22} \approx 0.098, which is small enough. With 22 calls the chance of getting through is about 1 - 0.098 = 0.902.

Figure 1The chance of at least one success in n tries, 1 - 0.9^n, when each try succeeds with probability 0.1. It climbs toward 1 but never reaches it, and first passes 0.9 at n = 22.

Figure 1 shows why so many calls are needed. The first few calls add a lot, 0.1 then 0.09 then 0.081, but each later call can only add a share of what is still missing, so the curve flattens as it nears 1.

Example 5

Eggs taken from a crate

A crate holds 15 eggs, and 4 of them are cracked. A chef takes 5 eggs, one at a time, without putting any back. Find the probability that at least one of the five is cracked. Compare it with the answer if each egg were put back before the next was taken.

Solution

The draws are not independent: each good egg taken leaves fewer good eggs behind. “None cracked” is still a single path, good, good, good, good, good, and its chance is a product of updated fractions. There are 11 good eggs of 15 at the first draw, then 10 of 14, 9 of 13, 8 of 12 and 7 of 11:

\begin{aligned} P(\text{none cracked}) &= \frac{11}{15} \times \frac{10}{14} \times \frac{9}{13}\\ &\quad \times \frac{8}{12} \times \frac{7}{11}\\ &= \frac{55440}{360360} = \frac{2}{13} \end{aligned}

P(\text{at least one cracked}) = 1 - \frac{2}{13} = \frac{11}{13} \approx 0.846

If each egg went back, every draw would be good with probability \displaystyle \frac{11}{15}, independently, so

\begin{aligned} P(\text{at least one cracked}) &= 1 - \left(\frac{11}{15}\right)^5\\ &= 1 - \frac{161051}{759375} \approx 0.788 \end{aligned}

Keeping the eggs out makes a cracked one more likely to turn up, because every good egg removed raises the cracked eggs’ share of what is left. The complement trick works the same way in both cases; only the factors in the product change.

Example 6

Sharing a birth month

Five people meet. Treat each person’s birth month as equally likely to be any of the 12 months, independently of the others. Find the probability that at least two of them share a birth month.

Solution

“At least two share a month” means at least one pair matches, which covers a great many arrangements. Its complement is “all five months are different”. Go through the people in turn. The first can have any month. The second must avoid that one month: 11 of 12. The third must avoid two months: 10 of 12. Then 9 of 12 and 8 of 12:

\begin{aligned} P(\text{all different}) &= \frac{12 \times 11 \times 10 \times 9 \times 8}{12^5}\\[2pt] &= \frac{95040}{248832} = \frac{55}{144} \end{aligned}

P(\text{at least two share}) = 1 - \frac{55}{144} = \frac{89}{144} \approx 0.618

With only five people and twelve months, a shared month is more likely than not. Each factor here is a conditional chance, since what the third person must avoid depends on the first two being different; this is the without-replacement pattern of the eggs, applied to months.

Boundaries

The product of failures needs independence. The formula 1 - (1 - p)^n is true only when the tries do not affect each other. For draws kept out, or any tries that influence one another, the product must use updated chances, as with the eggs. Treating linked events as independent gives a wrong number, not a rough one.

The complement must be the exact negation. The complement of “at least one success” is “no successes”, and the complement of “at least one failure” is “all successes”. The complement of “at least two” is “none or exactly one”, so “at least two” needs both of those subtracted from 1.

At least one is never certain when each try can fail. If 0 < p < 1, then (1 - p)^n is positive for every n, so 1 - (1 - p)^n is always less than 1. More tries bring it as close to 1 as you like, but no finite number of tries reaches it. If p = 0 the chance stays 0 forever.

The tries-needed inequality turns over. Dividing by \log(1 - p) reverses the inequality because that logarithm is negative whenever 0 < p < 1. The answer is then rounded up, never down, since the inequality asks for at least that many tries.

The model is part of the answer. Equal and independent birth months, or the same chance of getting through on every call, are assumptions. The arithmetic is exact for the model; how well the model fits the world is a separate question.

Where this goes next

The expression 1 - (1 - p)^n is the chance that the first success comes somewhere in the first n tries, which is the cumulative side of the geometric distribution: the first success lands exactly on try k with probability (1 - p)^{k - 1}p. Asking for exactly k successes in n tries, rather than at least one, leads to the binomial probability formula and then the binomial distribution, where “none” is the k = 0 case and “at least one” is everything else. For draws without replacement from a large group, the products of updated fractions are organized by probability with counting formulas.

Exercises

Equal chances, independent tries

A fair coin is tossed 7 times. Find the probability of at least one tail.
A fair number cube is rolled 3 times. Find the probability that at least one roll is a 1.
A quality check rejects each part with probability 0.05, independently of the other parts. Four parts are checked. Find the probability that at least one is rejected.
Eight plant cuttings each take root with probability 0.7, independently. Find the probability that (a) at least one takes root, (b) at least one fails to take root.

Different chances

Three friends each try to buy a concert ticket online, succeeding independently with probabilities 0.6, 0.5 and 0.25. Find the probability that at least one of them gets a ticket.
Two fire lookouts each watch the same forest. They spot a new fire independently, with probabilities 0.7 and 0.65. Find the probability that the fire is spotted.
A notice is sent by email, by text and by letter, which arrive independently with probabilities 0.98, 0.95 and 0.9. Find the probability that it reaches its reader at least once.

Drawing without replacement

A bag holds 11 marbles, 4 of them green. Three marbles are taken, one at a time, and kept out. Find the probability that at least one is green.
In a group of 14 students, 5 wear glasses. Two different students are picked at random. Find the probability that at least one wears glasses. Then find it if the same student could be picked twice.
Four cards are dealt from a shuffled standard deck of 52 cards, which holds 4 aces. Find the probability that at least one ace is dealt.

How many tries

How many times must a fair number cube be rolled for the probability of at least one six to be at least 0.95?
Each text in a survey gets a reply with probability 0.2, independently. How many texts must be sent for the probability of at least one reply to be at least 0.99?
Three independent tries each succeed with the same probability p, and the probability of at least one success is 0.973. Find p.
An event has a 0.001 chance of happening on any given day, independently from day to day. After how many days is the probability that it has happened at least once first at least 0.5?

Beyond at least one

A fair number cube is rolled 4 times. Find the probability of at least two sixes.
Four people meet. With birth months equally likely and independent, find the probability that at least two share a birth month.
A fair number cube is rolled 3 times. Find the probability of at least one six and at least one 1.
A machine needs a part that works with probability 0.9. Design A uses two such parts and runs only if both work. Design B uses two such parts and runs if at least one works. The parts work independently. Find the probability that each design runs.

Reasoning

Explain why the complement of “at least one of the tries succeeds” is “every try fails”, and write the complement of “at least one of the tries fails”.
Show that, for 0 < p < 1, the probability 1 - (1 - p)^n grows each time n goes up by one, and explain why it never reaches 1.
Each try succeeds with probability 0.1. Someone says ten tries make a success certain, since 10 \times 0.1 = 1. Find the real probability of at least one success in 10 tries and explain what the product 10 \times 0.1 does measure.
Explain why, in Example 5, keeping the eggs out made at least one cracked egg more likely than putting them back. When would the two answers be almost the same?

Answers

\displaystyle \frac{127}{128}. No tail means seven heads, \displaystyle \left(\frac{1}{2}\right)^7 = \frac{1}{128}, and \displaystyle 1 - \frac{1}{128} = \frac{127}{128}.
\displaystyle \frac{91}{216}. No 1 in three rolls is \displaystyle \left(\frac{5}{6}\right)^3 = \frac{125}{216}, so the answer is \displaystyle 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421.
About 0.185. None rejected is 0.95^4 = 0.81450625, and 1 - 0.81450625 = 0.18549375.
(a) 1 - 0.3^8 = 0.99993439 \approx 0.99993, since none taking root is 0.3^8. (b) 1 - 0.7^8 = 0.94235199 \approx 0.942, since the complement is all eight taking root.
0.85. All three fail with probability 0.4 \times 0.5 \times 0.75 = 0.15, and 1 - 0.15 = 0.85.
0.895. Both miss with probability 0.3 \times 0.35 = 0.105, and 1 - 0.105 = 0.895.
0.9999. All three fail with probability 0.02 \times 0.05 \times 0.1 = 0.0001, and 1 - 0.0001 = 0.9999.
\displaystyle \frac{26}{33} \approx 0.788. No green is \displaystyle \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9} = \frac{210}{990} = \frac{7}{33}, and \displaystyle 1 - \frac{7}{33} = \frac{26}{33}.
\displaystyle \frac{55}{91} \approx 0.604, and \displaystyle \frac{115}{196} \approx 0.587 with repeats allowed. Neither in glasses is \displaystyle \frac{9}{14} \times \frac{8}{13} = \frac{36}{91}; with repeats it is \displaystyle \left(\frac{9}{14}\right)^2 = \frac{81}{196}.
\displaystyle \frac{15229}{54145} \approx 0.281. No ace is \displaystyle \frac{48}{52} \times \frac{47}{51} \times \frac{46}{50} \times \frac{45}{49} = \frac{38916}{54145}, and \displaystyle 1 - \frac{38916}{54145} = \frac{15229}{54145}.
17 rolls. Need \displaystyle \left(\frac{5}{6}\right)^n \le 0.05, so \displaystyle n \ge \frac{\log 0.05}{\log \frac{5}{6}} \approx 16.43. Check: \displaystyle \left(\frac{5}{6}\right)^{16} \approx 0.054 is too big and \displaystyle \left(\frac{5}{6}\right)^{17} \approx 0.045 is not.
21 texts. Need 0.8^n \le 0.01, so \displaystyle n \ge \frac{\log 0.01}{\log 0.8} \approx 20.64. Check: 0.8^{20} \approx 0.0115 and 0.8^{21} \approx 0.0092.
p = 0.7. The chance of no success is (1 - p)^3 = 1 - 0.973 = 0.027, and the only real cube root of 0.027 is 0.3, so 1 - p = 0.3.
693 days. Need 0.999^n \le 0.5, so \displaystyle n \ge \frac{\log 0.5}{\log 0.999} \approx 692.8. Check: 0.999^{692} \approx 0.5004 and 0.999^{693} \approx 0.4999.
\displaystyle \frac{19}{144} \approx 0.132. The complement is none or exactly one six. None is \displaystyle \frac{625}{1296}. Exactly one is 4 positions times \displaystyle \frac{1}{6} \times \left(\frac{5}{6}\right)^3, which is \displaystyle \frac{500}{1296}. So the answer is \displaystyle 1 - \frac{1125}{1296} = \frac{171}{1296} = \frac{19}{144}.
\displaystyle \frac{41}{96} \approx 0.427. All different is \displaystyle \frac{12 \times 11 \times 10 \times 9}{12^4} = \frac{11880}{20736} = \frac{55}{96}, and \displaystyle 1 - \frac{55}{96} = \frac{41}{96}.
\displaystyle \frac{5}{36}. The complement is “no six or no 1”. No six and no 1 each have probability \displaystyle \left(\frac{5}{6}\right)^3 = \frac{125}{216}, and both together (every roll from 2 to 5) has \displaystyle \left(\frac{4}{6}\right)^3 = \frac{64}{216}. These overlap, so by the general addition rule the complement is \displaystyle \frac{125 + 125 - 64}{216} = \frac{186}{216}, and the answer is \displaystyle 1 - \frac{186}{216} = \frac{30}{216} = \frac{5}{36}.
Design A 0.81, design B 0.99. A needs both: 0.9 \times 0.9 = 0.81. B fails only if both fail: 0.1 \times 0.1 = 0.01, so it runs with probability 1 - 0.01 = 0.99.
Each outcome has some number of successes. “At least one succeeds” holds every outcome with 1 or more; the only outcomes left have 0 successes, which is every try failing. The two events share no outcome and together cover all of them, so they are complements. The complement of “at least one fails” is “every try succeeds”.
Going from n to n + 1 tries changes (1 - p)^n to (1 - p)^n(1 - p), which is smaller because (1 - p)^n > 0 and 0 < 1 - p < 1. A smaller amount subtracted from 1 leaves more, so 1 - (1 - p)^n grows by (1 - p)^n p > 0. It never reaches 1 because (1 - p)^n, a product of positive numbers, is never 0.
About 0.651. No success in ten tries is 0.9^{10} \approx 0.349, so at least one is 1 - 0.9^{10} \approx 0.651, far from certain. The product 10 \times 0.1 = 1 is the number of successes to expect on average over many runs of ten tries; some runs have none and some have two or more, which balances out.
Each good egg taken out leaves the same 4 cracked eggs in a smaller crate, so the next draw is more likely to be cracked than \displaystyle \frac{4}{15}. With replacement the share stays \displaystyle \frac{4}{15} every time. The two answers are almost the same when the crate is large compared with the number of eggs taken, because then removing a few barely changes the shares.