Probability of Permutation Events | Free Discrete Mathematics Lesson (4 College Credits) | UPI Study

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Probability of Permutation Events

Discrete Mathematics · Chapter 5: Counting Methods and Basic ProbabilitySection 5.5 · lesson 31 of 879 min · free with a free account

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In this lesson

Count every arrangement, count the ones the event allows, divide.

  • Turn a probability question into 2 permutation counts
  • Count favorable arrangements by fixing positions
  • Handle adjacency with the block technique
  • Use the complement when exclusion is easier

Transcript

Everything said in the video, section by section (1,275 words). Read along, or search it.

Counting Methods and Basic Probability

Probability of Permutation Events

This section combines two tools already in hand. Permutation counts from section five point four, and classical probability from section five point one. Whenever the sample space is a set of equally likely arrangements, a probability question becomes two counting questions. Count every possible arrangement. Then count only the arrangements the event allows. And divide.

The Opening Problem

Where Do the Decoys Land?

An escape-room company resets one of its puzzles between groups. It shuffles five numbered keycards, and hangs them on five hooks in a random order. That way, every group faces a genuinely new layout. The design team wants to know something specific. Across all the random layouts, what fraction put the two decoy keycards on the two hooks closest to the door? Answering that needs no new formula. It's a permutation count, answered twice. Once for every possible arrangement, and once for only the arrangements the event allows.

5.5.1 The Framework

Probability as a Ratio of Arrangement Counts

Here's the framework. Suppose the sample space is every possible arrangement of some items. A random seating, a random finish order, or a randomly generated code. And suppose every arrangement is equally likely. Then the probability of an event is the number of arrangements satisfying it, divided by the total number of arrangements. The total is usually n P r, or n factorial when every item is being arranged. The favorable count comes from applying the multiplication rule to whichever positions the event restricts. Then multiply by the permutations of whatever's left unrestricted. The examples ahead apply that same recipe to seating, rankings, and passwords. The arithmetic changes with the scenario, but the structure never does.

Table 5.5.1

The Permutation-Probability Recipe

This table lays out the recipe as four explicit steps. Step one. Count the total, every possible ordering of the sample space. That's n factorial, or n P r when arranging r of n items. Step two. State precisely what an arrangement must satisfy to count as favorable. Fixed positions, a subset of positions, two items next to each other, or an excluded item. Step three. Count the favorable arrangements. Handle the restricted positions first, then multiply by the permutations of everything else. And step four. Divide the favorable count by the total count.

Worked Example 5.5.1

2 Specific People at the Ends

First, seating. Five coworkers, Alice, Bob, Carol, Dave, and Eve, sit in a row of five numbered seats, in a uniformly random order. What's the probability that Alice and Bob end up in the two end seats, seats one and five, in either order? The total is all five people in five seats. That's five factorial, one hundred twenty. For the favorable count, Alice and Bob take the two end seats, in two possible orders. Then the other three fill the three middle seats freely, in three factorial, or six ways. By the multiplication rule, that's two times six, twelve favorable arrangements. So the probability is twelve over one hundred twenty, which is one tenth.

Worked Example 5.5.2

1 Specific Person in the Middle

Same five coworkers, same five seats. What's the probability that Carol specifically ends up in the middle seat, seat three? The total is still one hundred twenty. For the favorable count, Carol is fixed in seat three, one way. The other four coworkers fill the other four seats in any order. That's four factorial, twenty-four ways. So the probability is twenty-four over one hundred twenty, which is one fifth. And the escape-room question has exactly the structure of the previous example. Two specific keycards on two specific hooks, in either order, with three cards free. So its answer is also two times three factorial, over five factorial. One tenth of all layouts.

Worked Example 5.5.3

Finishing in the Top 3

Next, rankings. Eight sprinters run a final, and every one of the eight factorial finish orders is equally likely. What's the probability that a particular sprinter, Priya, finishes in the top three? The total is eight factorial, forty thousand, three hundred twenty. For the favorable count, Priya takes one of the three top positions. That's three ways. The other seven sprinters fill the other seven positions freely, in seven factorial ways. That's three times five thousand forty, or fifteen thousand, one hundred twenty. Divided by the total, it's three eighths. That matches the intuitive shortcut. In a uniformly random finish, Priya is equally likely to land in any of the eight positions, and three of them are top three. The counting method reaches the same answer, without relying on that shortcut.

Worked Example 5.5.4

2 Finishers Side by Side

A different race has six runners, with all six factorial finish orders equally likely. That's seven hundred twenty. What's the probability that two particular runners, X and Y, finish in adjacent positions, in either order? Here's the trick. Temporarily glue X and Y together into a single block. Arranging that block with the other four runners means arranging five units. That's five factorial, one hundred twenty ways. Then split the block apart again. Inside it, X and Y can appear in two orders, X then Y, or Y then X. So there are one hundred twenty times two, which is two hundred forty favorable arrangements. And the probability is two hundred forty over seven hundred twenty, one third. Treating two items as one block is the standard way to count an adjacency requirement.

Worked Example 5.5.5

Fixed First and Last Characters

Now passwords. An access system generates a password by arranging five of eight available symbols, with no repetition. Every possible password is equally likely. What's the probability that a random password starts with the at sign, and ends with the question mark? The total is eight P five. Eight times seven times six times five times four, which is six thousand, seven hundred twenty. For the favorable count, the first position is fixed as the at sign, and the last as the question mark. One way each. The middle three positions take any three of the six remaining symbols, in order. That's six P three, one hundred twenty ways. So the probability is one hundred twenty over six thousand, seven hundred twenty, which is one fifty-sixth.

Worked Example 5.5.6

Excluding a Symbol - and the Complement

A different system arranges five of nine symbols, again with no repetition. What's the probability that the password doesn't include the hash symbol anywhere? The total is nine P five, fifteen thousand, one hundred twenty. If the hash can't appear at all, every position is filled from the other eight symbols. That's eight P five, six thousand, seven hundred twenty. So the probability that the hash is absent is four ninths. Now the complement. The arrangements where an event doesn't happen number the total, minus the ones where it does. So the probability of not E is one, minus the probability of E. So the probability that the hash appears somewhere in the password is one minus four ninths, which is five ninths. That's often the faster route, whenever the complementary count is easier to build than the direct one.

Takeaways

2 Counts and a Division

To summarize. When a sample space consists of equally likely arrangements, the probability of an event is the favorable arrangements over the total. Both come from the multiplication rule and n P r. Fixing items at specific positions reduces the favorable count to a smaller permutation problem, over whatever's left. Requiring two items to be adjacent is counted by treating them as a single block, and multiplying by the block's internal orders. And the complement rule pairs naturally with counting, whenever excluding an item is easier to count than including it. Here's something to think about. Suppose a problem asks for the probability that a random three-person committee, chosen from eight people, includes a specific person. A committee has no internal order. But what if both the favorable and total outcomes were counted with ordering, as with eight P three? Would the ratio still be right? Think about what role order actually plays, when the numerator and denominator are overcounted in exactly the same way.

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FAQ

Questions students ask us

About Probability of Permutation Events

How do you find the probability of a permutation event?
Divide the number of favorable arrangements by the total number of arrangements, assuming every arrangement is equally likely. The total is n! when all items are arranged, or _nP_r when arranging r of n. Count the favorable arrangements by handling restricted positions first, then multiplying by the permutations of the remaining items.
How do you find the probability that two specific people sit in the end seats of a row of five?
The probability is \frac{2\cdot 3!}{5!}=\frac{1}{10}. The total is 5!=120 seatings. For favorable ones, the two people fill seats one and five in 2 orders, and the other three people fill the middle seats in 3!=6 ways. That gives 2\cdot 6=12 favorable arrangements, and 12/120=1/10.
How do you find the probability that two items are next to each other in a random arrangement?
Treat the two items as one block, count arrangements of the blocks, then multiply by the block's internal orders. For 6 runners, the chance that X and Y finish adjacent is found by arranging 5 units in 5!=120 ways. The block has 2 internal orders, so 240/720=1/3.
When should you use the complement to find a permutation probability?
Use the complement when counting the outcomes where the event does not happen is easier than counting where it does. Then P(E)=1-P(\text{not }E). For a 5-symbol password from 9 symbols, the chance that a hash is absent is \frac{_8P_5}{_9P_5}=\frac{4}{9}. So the chance it appears is 1-\frac{4}{9}=\frac{5}{9}.

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