Section PC.8 – Probability with Permutations and Combinations
Up to this point we have been using permutations to help us count, but we can also use permutations to help us answer more complex probability questions. In the following examples, we will use permutations to find probabilities of certain events.
| Example 1 |
| A 4 digit PIN number is selected. What is the probability that there are no repeated digits? |
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There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 · 10 · 10 · 10 = 104 = 10000 total possible PIN numbers. To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 · 9 · 8 · 7, or notice that this is the same as the permutation 10P4 = 5040. The probability of no repeated digits is the number of 4 digit PIN numbers with no repeated digits divided by the total number of 4 digit PIN numbers. This probability is: [latex]\frac{_{10}P_4}{10^4}[/latex] = [latex]\frac{5,040}{10,000}[/latex] = [latex]0.504[/latex] |
| Birthday Problem |
| Let’s take a pause to consider a famous problem in probability theory:
Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday? Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer ([latex]\frac{30}{365}[/latex], perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however. |
| Example 2 |
| Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people? |
| There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are no shared birthdays. In other words, the alternative to “at least one” is having none. In other words, since this is a complementary event,
P(at least one) = 1 – P(none) We will start, then, by computing the probability that there is no shared birthday. Let’s imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday? There are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is [latex]\frac{364}{365}[/latex]. Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the third person does not share a birthday with the first two is [latex]\frac{363}{365}[/latex]. We want the second person not to share a birthday with you and the third person not to share a birthday with the first two people, so we use the multiplication rule: P(no shared birthday) = [latex]\frac{365}{365}[/latex] · [latex]\frac{364}{365}[/latex] · [latex]\frac{363}{365}[/latex] ≈ [latex]0.9918[/latex] and then subtract 1 to get: P(shared birthday) = 1 – P (no shared birthday) = [latex]1[/latex] – [latex]0.9918[/latex] = [latex]0.0082[/latex] |
This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let’s make our group a bit bigger.
| Example 3 |
| Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people? |
| Continuing the pattern of the previous example, the answer should be:
P(shared birthday) = [latex]1[/latex] – [latex]\frac{365}{365}[/latex] · [latex]\frac{364}{365}[/latex] · [latex]\frac{363}{365}[/latex] · [latex]\frac{362}{365}[/latex] · [latex]\frac{361}{365}[/latex] ≈ [latex]0.0271[/latex] Note that we could rewrite this more compactly as: P(shared birthday) = [latex]1[/latex] – [latex]\frac{_{365}P_5}{365^5}[/latex] ≈ [latex]0.0271[/latex] which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group. |
| Example 4 |
| Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people? |
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Here we can calculate: P(shared birthday) = [latex]1[/latex] – [latex]\frac{_{365}P_30}{365^30}[/latex] ≈ [latex]0.706[/latex] which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday! |
If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.
| You Try PC.8.A |
| Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people? |
Just as we did with permutations, we can now use combinations to move beyond counting and help us answer more complex probability questions.
| Example 5 |
| In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket. |
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In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is 48C6 = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is: [latex]\frac{_{6}C_6}{_{48}C_6}[/latex] = [latex]\frac{1}{12,271,512}[/latex] ≈ [latex]0.0000000815[/latex] |
| Example 6 |
| In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket. |
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As above, the number of possible outcomes of the lottery drawing is 48C6 = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by 6C5 = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by 42C1 = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: 6C5 · 42C1 = 6 · 42 = 252. So the probability of winning the second prize is: [latex]\frac{(_6C_5)(_{42}C_1)}{_{48}C_6}[/latex] = [latex]\frac{252}{12,271,512}[/latex] ≈ [latex]0.0000205[/latex] |
| Example 7 |
| Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace. |
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In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus, we use combinations to compute the possible number of 5-card hands, 52C5. This number will go in the denominator of our probability formula, since it is the number of possible outcomes. For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be 4C1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48C4 ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be 4C1 · 48C4 ways to choose one ace and four non-Aces. Putting this all together, we have: P(one Ace) = [latex]\frac{(_4C_1)(_{48}C_4)}{_{52}C_5}[/latex] = [latex]\frac{778,320}{2,598,960}[/latex] ≈ [latex]0.299[/latex] |
| Example 8 |
| Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces. |
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The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same: P(two Aces) = [latex]\frac{(_4C_2)(_{48}C_3)}{_{52}C_5}[/latex] = [latex]\frac{103,776}{2,598,960}[/latex] ≈ [latex]0.0399[/latex] |
| You Try PC.8.B |
| Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings. |
Section PC.8 Answers to You Try Problems
PC.8.A
0.1169
PC.8.B
0.000009234
