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

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Probability of Combination Events

Discrete Mathematics · Chapter 5: Counting Methods and Basic ProbabilitySection 5.8 · lesson 34 of 877 min · free with a free account

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In this lesson

Counting subsets on top and bottom of a probability.

  • Apply a 4-step method for combination probabilities
  • Count selections drawn from separate groups
  • Compute partial-match lottery probabilities
  • See 1 calculation behind different-looking problems

Transcript

Everything said in the video, section by section (1,014 words). Read along, or search it.

Counting Methods and Basic Probability

Probability of Combination Events

This section closes chapter five by joining two ideas already built. Classical probability, and combinations. Whenever a selection is made without regard to order, and every selection is equally likely, a probability comes down to counting subsets, on top and on bottom. We'll work through three of the most common settings. Card hands, committees, and lottery draws.

The Opening Problem

Random, Order-Independent Samples

Software systems draw random, order-independent samples all the time. An audit tool selects a handful of transactions to review, out of a much larger log. A testing framework samples a subset of user accounts, for a staged rollout. Whenever every such selection is equally likely, the classical probability formula from section five point one applies directly. The probability of an event is the size of the event, over the size of the sample space. And computing it means counting subsets, which is exactly what the combination formula from section five point seven does. Here it's written C of n comma r.

The General Method

The Same 4 Steps Every Time

Using combinations inside a probability calculation follows the same four steps every time. First, identify the sample space, the set of all equally likely, order-independent selections. Second, identify the event, the subset of those selections that satisfy the condition you care about. Third, count both, using combinations. And fourth, divide. The combination formula itself isn't re-derived here. Section five point seven is taken as given.

Worked Example 1

A 5-Card Hand, All Hearts

First, a card hand. Five cards are dealt at random from a standard fifty-two card deck. What's the probability that all five are hearts? The sample space is every possible five-card hand, all equally likely. Order doesn't matter in a hand, so its size is C of fifty-two, five. That's two million, five hundred ninety-eight thousand, nine hundred sixty hands. The event is the hands using only hearts. There are thirteen hearts, and a favorable hand chooses all five cards from just those thirteen. That's C of thirteen, five, which is one thousand two hundred eighty-seven. Dividing gives about zero point zero zero zero four nine five. Roughly a five hundredths of a percent chance.

Figure 1

Only the Hearts Row Can Supply the Hand

Here's the whole deck, laid out as four rows of thirteen ranks. An all-hearts hand can only be assembled from the thirteen highlighted cards, in the hearts row. That's exactly why the favorable count chooses five from thirteen, and not five from fifty-two.

Worked Example 2

A Committee From 2 Groups

Many selections draw from more than one group at once. A committee might come partly from one pool of people, and partly from another, with a set number from each. Here's an example. A five-person committee is chosen at random from eight engineers and five designers, thirteen people in all. What's the probability it has exactly three engineers, and two designers? The sample space is every five-person committee from thirteen. That's C of thirteen, five, one thousand two hundred eighty-seven. A favorable committee chooses three of the eight engineers, in C of eight, three, fifty-six ways. And independently, two of the five designers, in C of five, two, ten ways. Each engineer choice pairs with each designer choice, so multiply. That's five hundred sixty favorable committees. So the probability is five hundred sixty over one thousand two hundred eighty-seven, about zero point four three five one. Roughly forty-three and a half percent.

Checking in Code

Both Examples, Built-In comb()

Python's built-in comb function checks both examples. C of thirteen, five, over C of fifty-two, five, rounds to zero point zero zero zero four nine five. And C of eight, three, times C of five, two, over C of thirteen, five, rounds to zero point four three five one. Both match the hand calculations.

Worked Example 3

Matching Exactly 4 Lottery Numbers

A lottery draw splits a pool of numbers the same way a committee splits a pool of people. The numbers that were drawn, and the numbers that weren't. Here's the problem. A lottery draws six numbers from forty-nine. A player buys one ticket of six numbers, chosen in advance. What's the probability of matching exactly four? The sample space is every set of six numbers from forty-nine. That's C of forty-nine, six, which is thirteen million, nine hundred eighty-three thousand, eight hundred sixteen. Now split the forty-nine into the six drawn and the forty-three not drawn. A ticket with exactly four matches has four of the drawn numbers, and two of the others. Both parts are required, since the ticket has exactly six numbers. That's C of six, four, fifteen, times C of forty-three, two, nine hundred three. So there are thirteen thousand five hundred forty-five favorable tickets, for a probability of about zero point zero zero zero nine six nine. And the jackpot is a special case. Matching all six means choosing all six drawn numbers, and none of the others. That's exactly one ticket, out of nearly fourteen million.

The Same Pattern, 3 Settings

People, Numbers - One Calculation

Notice the structure across all three settings. The card hand chose all five cards from a single thirteen-card suit. The committee split a pool of thirteen people into eight and five, chose from each part, and multiplied. The lottery split a pool of forty-nine numbers into six and forty-three, chose from each part, and multiplied in exactly the same way. The problems look unrelated on the surface. One is about people, the other about numbers. But they're the same calculation, applied to different labels.

Takeaways

Split, Choose, Multiply, Divide

To summarize. Using combinations in a probability always follows the same pattern. Identify the sample space and the event, count each with combinations, and divide. When a selection draws a required number from each of several groups, multiply the separate combination counts. Any valid choice from one group pairs with any from another. Committees and lottery partial matches use the identical split and multiply technique. And a jackpot is the special case where every drawn item is matched. Here's something to think about. A student solves the all-hearts example with permutations throughout, top and bottom. The answer matches exactly. They conclude it doesn't matter which you use, as long as you're consistent. Is that right in general? Why did the permutations cancel in the card example? And for the committee, would eight P three times five P two, over thirteen P five, still give zero point four three five one? What's different?

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FAQ

Questions students ask us

About Probability of Combination Events

What are the steps for finding a probability with combinations?
Use four steps: identify the sample space, identify the event, count both with combinations, and divide. The sample space is all equally likely, order-independent selections, and the event is those meeting the condition. For all hearts in a 5-card hand, \binom{13}{5}/\binom{52}{5} = 1287/2598960 \approx 0.000495.
How do you find the probability of drawing a required number from each of several groups?
Count the ways to choose from each group with combinations, multiply those counts, and divide by the total number of selections. For a 5-person committee from 8 engineers and 5 designers, exactly 3 engineers and 2 designers gives \binom{8}{3}\binom{5}{2}/\binom{13}{5} = 56\cdot 10/1287 = 560/1287 \approx 0.4351.
How do you calculate the probability of matching exactly 4 numbers in a 6-of-49 lottery?
Split the 49 numbers into the 6 drawn and the 43 not drawn, choose 4 from the first group and 2 from the second, then divide by all tickets. This gives \binom{6}{4}\binom{43}{2}/\binom{49}{6} = 15\cdot 903/13983816 = 13545/13983816 \approx 0.000969.
Can I use permutations instead of combinations in a probability problem?
Not safely in general. Permutations in both the numerator and denominator cancel for all hearts, since both count orderings the same way. For the committee, \frac{{}_8P_3\cdot{}_5P_2}{{}_{13}P_5} = 6720/154440 \approx 0.0435, not 0.4351. That product misses how the roles can be interleaved in order, so combinations are the reliable choice.

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