Use the Fundamental Counting Principle , permutations, and combinations to find probability. To compute theoretical probability using the formula \(P(E)=\frac{n(E)}{n(S)\), we must know how many outcomes are in the sample space \(S\) and how many outcomes belong to the event \(E\). When these numbers are small, we can list and count the outcomes directly. However, many real‑world situations involve far too many outcomes to list, so we rely on counting techniques to determine these quantities efficiently. In this section, you will apply what you have learned about permutations, combinations, and the Fundamental Counting Principle to count the number of possible outcomes in a sample space or event. Once these counts are known, we will use them in the probability formula to compute theoretical probabilities accurately. Probability with the Fundamental Counting Principle In finding any probability, we follow the same three steps: First, count the outcomes in the sample space, \(n(S)\). Next, count the favorable outcomes in the event, \(n(E)\). Finally, substitute these counts into the probability formula, \(P(E)=\dfrac{n(E)}{n(S)}\). The next few examples demonstrate how the Fundamental Counting Principle allows us to quickly determine \(n(S)\) and \(n(E)\). Each question on a 5‑question multiple‑choice quiz has four possible answer choices: A, B, C, and D. Randy randomly guesses on every question. What is the probability that he gets all a perfect score (all five questions correct)? First, count the total number of possible answer patterns. Each question has 4 possible choices, and there are 5 questions. By the Fundamental Counting Principle the total number of possible answer patterns is \(4 \times 4 \times 4 \times 4 \times 4=4^5=1024.\) There are 1,024 different ways Randy could answer the quiz. This corresponds to the sample space so \(n(S)=1,024\). Next, count the number of favorable answer patterns for the event "all 5 questions correct." Randy must choose the one correct answer for each of the 5 questions. There is exactly 1 way to do this: the unique pattern where every answer matches the key. There is 1 way that Randy could answer all the questions correctly for a perfect score. This corresponds to the event so \(n(E)=1\). Finally, calculate the probability. \(P(E)= \dfrac{n(E)}{n(S)} = \dfrac{1}{1024}\). So the chance that Randy gets a perfect score by random guessing is \(\frac{1}{1024} \approx 0.00098 \) or about 0.098%. The letters in the word HOLIDAY are randomly scrambled. What is the probability that the first two letters in the arrangement are both vowels? First, count the number of ways that the 7 letters could be scrambled. Note that that the letters in HOLIDAY are distinct. By the Fundamental Counting Principle, the number of possible permutations is \(7!=7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=5,040\). This corresponds to the sample space so \(n(S)=5,040\). Next, count the arrangements where the first two letters are vowels. The vowels in HOLIDAY are O, I, A -- three vowels total. There are 3 choices for the first letter (any vowel). After choosing one vowel, 2 choices remain for the second letter. Once the first two positions are filled with vowels, the 5 remaining letters can be arranged in \(5!\) ways. Using the Fundamental Counting Principle, there are \( 3 \cdot 2 \cdot 5!=720\) favorable arrangements for this event. So, \(n(E)=720\). Finally, calculate the probability. \(P(E)= \dfrac{n(E)}{n(S)} = \dfrac{720}{5040} = \dfrac{1}{7}\). The probability that the first two letters in the arrangement are both vowels is \(\frac{1}{7} \approx 0.1429 \text{ or about 14.29\%}\). A locker combination consists of 7 distinct digits arranged in a random order. The digits available are 0, 1, 2, 3, 4, 5, 6. What is the probability that the first three digits in the combination are even numbers? \(\frac{576}{5,040}=\frac{4}{35} \approx 0.1143 \approx 11.43\%\) Banks often assign a 4‑digit PIN (personal identification number) for debit cards. Assume these PINs are generated at random. What is the probability that a randomly assigned PIN has no repeated digits? First, count all possible 4‑digit PINs. Each digit can be any of the ten digits 0–9. By the Fundamental Counting Principle, the number of PINS possible is \(10 \cdot 10 \cdot 10 \cdot 10 =10,000\). There are 10,000 possible PINS (from 0000 to 9999). This corresponds to the sample space so \(n(S)=10,000\). Next, count the PINs with no repeated digits. To form a PIN so that no digit is repeated, there are 10 choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit. Using the Fundamental Counting Principle, there are \( 10 \cdot 9 \cdot 8 \cdot 7 =5,040\) favorable arrangements for this event: \(n(E)=5,040\). Finally, calculate the probability. \(P(\text{no repeated digits})= \dfrac{n(E)}{n(S)} = \dfrac{5,040}{10,000} = 0.504 = 50.4\%\). A security system generates a 3-character access code using the letters A–Z (uppercase only). Each character is chosen at random. What is the probability that the access code contains no repeated letters? \(\frac{15,600}{17,576}=\frac{150}{169}=0.8875=88.75\%\) Probability with Permutations and Combinations As we continue exploring probability, we now turn to situations where the order of selection matters or where we are forming groups without repetition. In these cases, the Fundamental Counting Principle still underlies our reasoning, but the counting becomes more structured. This is where permutations and combinations become essential tools. Permutations and combinations apply only when the objects being arranged or selected cannot be repeated. If changing the order produces a different outcome, we use a permutation. If the order does not affect the outcome, we use a combination. The next set of examples will show how these formulas allow us to compute probabilities efficiently when order and non‑repetition play a central role. In these problems, the same three steps will be used: Find \(n(S)\), next \(n(E)\), and finally use them to calculate \(P(E)\). A word game requires players to select 4 tiles from a bag containing 26 tiles, each labeled with one of the letters A through Z. What is the probability that a player will be able to spell CATS with the tiles he selects? First, count the number of possible groups of any 4 tiles that could be selected from the bag of 26 tiles. Since the order of the tiles does not matter when selecting a group of tiles, we evaluate the combination \(_{26}C_{4}=14,950\). There are 14,950 ways to select a group of any 4 tiles. This corresponds to the sample space so \(n(S)=14,950\). Next, count the favorable outcomes. To spell CATS , the player must select exactly the tiles C, A, T, S but the player can do so in any order. There is only one one way this can happen so \(n(E)=1\). If you are not convinced this is indeed only 1 way, we are choosing 4 tiles and they must be selected only from the tiles A, C, S, T: Finally, calculate the probability. \(P(E)) = \dfrac{n(E)}{n(S)} = \dfrac{1}{14,950}\). This probability that a player can spell CATS with the tiles is \(\frac{1}{14,950} \approx 0.000067 \text{ or } 0.0067\%\). This is a very small likelihood! Professor Jones’s desk contains 20 whiteboard markers, and 12 of them are depleted. He randomly selects five markers for class. What is the probability that all five markers selected are depleted? First, count the number of possible groups of any 5 markers that Professor Jones could select from the 20 markers in his desk. Because we are counting groups of markers, we evaluate the combination \(_{20}C_{5}=15,504\). There are 15,504 ways to select a group of any 5 markers. This corresponds to the sample space so \(n(S)=15,504\). Next, count the favorable outcomes. This event contains the outcomes where all 5 markers are depleted. They must be chosen from the 12 depleted markers in the desk. To count the number of ways this can occur evaluate \(_{12}C_5 = 792\). Finally, calculate the probability. \(P(E) = \dfrac{n(E)}{n(S)} = \dfrac{792}{15,504} = \dfrac{33}{646}\) The probability that all 5 markers are depleted is \(\frac{33}{646} \approx 0.0511 \text{ or }5.11\%\). A drawer contains 10 batteries, and 6 of them are fully charged. Three batteries are randomly selected for a device. What is the probability that all three batteries selected are fully charged? \(\frac{20}{120} = \frac{1}{6} \approx 0.1667 \text{ or }16.67\%\)

A school hosts an academic challenge featuring 8 competitors, and 4 of them are freshmen. All competitors have an equal chance of finishing in any position. What is the probability that freshmen finish in 1st, 2nd, and 3rd place? Because order matters in determining who finishes 1st, 2nd, and 3rd, this is a permutation probability problem. First, count the number of ways 1st, 2nd, and 3rd place can be awarded among the 8 competitors. Evaluate \(_8P_3=336\). This corresponds to the sample space so \(n(S)=336\). Next, count the favorable outcomes. Count the number of ways 1st, 2nd, and 3rd place can be awarded among the 4 freshmen. Evaluate \(_4C_3 = 24\) so \(n(E) = 24\). Finally, calculate the probability. \(P(E)) = \dfrac{n(E)}{n(S)} = \dfrac{24}{336} = \dfrac{1}{14}\). The probability that 1st, 2nd, and 3rd places are awarded to freshmen is \(\frac{1}{14} \approx 0.0714 \text{ or }7.14\%\). It should be noted here that probability problems involving a permutation can often be solved in more than one way if you think flexibly. In Example 7.3.7, we can also apply the Multiplicative Rule of Probability for dependent events. Suppose that \(F_1\) is the event that 1st place goes to a freshmen, \(F_2\) is the event that 2nd place goes to a freshmen, and \(F_3\) is the event that 3rd place goes to a freshmen. \(\begin{align*} P(F_1 \cap F_2 \cap F_3) &= P(F_1) \cdot P(F_2 \mid F_1) \cdot P(F_3 \mid F_1 \cap F_2) \\[6pt] &= \frac{4}{8} \cdot \frac{3}{7} \cdot \frac{2}{6} \\[6pt] &= \frac{24}{336} \\[6pt] &= \frac{1}{14} \end{align*}\) This matches the result obtained using permutations. A bag contains 10 colored marbles: 6 red and 4 yellow. You randomly select 5 marbles. What is the probability that all 5 marbles selected are red? Because order does not matter, we use combinations to count the number of possible groups of 5 marbles. First, count the number of ways of selecting 5 marbles from the bag. Evaluate \(_{10}C_5=252\). This corresponds to the sample space so \(n(S)=252\). Next, count the favorable outcomes. Count the number of ways to select 5 red marbles: \(n(E) = _6C_5 = 6\). Finally, calculate the probability. \(P(E) = \dfrac{n(E)}{n(S)} = \dfrac{6}{252} = \dfrac{3}{126}\) The probability that all five marbles are red is \(\frac{3}{126} = \approx 0.0238 \text{ or }2.38\%\). A box contains 12 colored pencils: 7 blue and 5 green. You randomly select 4 pencils. What is the probability that all 4 pencils selected are blue? \(\frac{35}{495} = \frac{7}{99} \approx 0.0707 \text{ or }7.07\%\). A bag contains 10 colored marbles: 6 red and 4 yellow. You randomly choose 5 marbles. What is the probability that exactly 3 of them are red? The sample space is the same as in Example 7.3.8. There are \(n(S)=252\) ways of selecting 5 marbles from the bag. To count the number of favorable outcomes in this event, we need to select 5 marbles so that 3 are red. This means the other 2 must be yellow. Use the Fundamental Counting Principle and find the product of two combinations. \(\begin{align*} {}_{6}C_{3} \times {}_{4}C_{2} &= 20 \times 6 \\[6pt] &= 120 \end{align*}\) So, \(n(E)=120\). Finally, calculate the probability. \(P(E) = \dfrac{n(E)}{n(S)} = \dfrac{120}{252} = \dfrac{10}{21}\) The probability that exactly 3 marbles are red is \(\frac{10}{21} \approx 0.4762 \text{ or }47.62\%\). Calculate the probability of being dealt exactly two hearts in a five‑card poker hand. A standard deck has 52 cards, including 13 hearts and 39 non‑hearts. Because order does not matter in a poker hand, we use combinations. First, count the number of ways of selecting 5 cards from the deck. Eveluate \(_{5}C_52=2,598,960\). This corresponds to the sample space so \(n(S)=2,598,960\). Next, count the favorable outcomes. Count the number of ways to select 2 hearts. But remember this means the other 3 cards are not hearts. Use the Fundamental Counting Principle and find the product of two combinations. \(\begin{align*} {}_{13}C_{2} \times {}_{39}C_{3} &= 78 \times 9,139 \\[6pt] &= 712,542 \end{align*}\) So, \(n(E)=712,542\). Finally, calculate the probability. \(P(E) = \dfrac{n(E)}{n(S)} = \dfrac{712,542}{2,598,960}\). The probability of selecting 2 hearts in a five-card poker hand is \(\frac{712,542}{2,598,960} \approx 0.2743 \text{ or }27.43\%\). Calculate the probability of being dealt four aces in a five-card poker hand. \(\frac{48}{2,598,960} \approx 0.000018 \text{ or }0.0018\%\) Section Summary Counting techniques allow us to analyze a wide variety of probability problems. In this section we used Fundamental Counting Principle, permutations, or combinations to count outcomes so we could apply the formula for theoretical probability. The choice depends on whether the situation involves sequential choices, ordered arrangements, or unordered selections. To find theoretical probability efficiently and effectively, every problem follows the same three steps: Count the outcomes in the sample space, \(n(S)\). Identify which outcomes satisfy the event and count the favorable outcomes in the event, \(n(E)\). Substitute these counts into the probability formula \(P(E)=\frac{n(E)}{n(S)}\). Section 7.3 Review Questions Four students with different last names randomly line up to go to lunch. What is the probability that they are lined up in alphabetical order by last name? What is the probability that a randomly assigned 5-digit student ID number has all odd digits? Assume digits may be repeated. The letters in the word NUMBER are arranged in random order. What is the probability that the first two letters are both consonants? A high school is staging the musical West Side Story . There are 20 singers auditioning for the musical. (We will assume they are equally talented.) The director is looking for two singers who could sing a good duet. What is the probability that Alicia and Juan are the two singers who are selected by the director? The director is also interested in the number of ways to choose a lead singer and a backup singer. What is the probability that Alicia is selected the lead singer and Juan is selected the backup singer? The director is looking for two singers who could sing a good duet. What is the probability that Alicia and Juan are the two singers who are selected by the director? The director is also interested in the number of ways to choose a lead singer and a backup singer. What is the probability that Alicia is selected the lead singer and Juan is selected the backup singer? A motorcycle license plate consists of 5 digits that are randomly selected. No digit is repeated. What is the probability of getting a license plate with all odd numbers? A technician is launching fireworks near the end of a show. Of the remaining 15 fireworks, 8 are blue, 4 are red, and 3 are white. If she launches 5 of them in a random order, what is the probability that she launches all blue fireworks? There are 12 males and 8 females in a jury pool. A group of 8 of them will be selected to sit on a jury. What's the probability of selecting 5 males and 3 females for the jury? There are 10 canned drinks in a cooler: 6 are colas and 4 are fruit drinks. Danielle will reach into the cooler and pull out 5 drinks without looking. Find the probability that all 5 drinks are colas. 3 drinks are colas and 2 drinks are fruit drinks. all 5 drinks are colas. 3 drinks are colas and 2 drinks are fruit drinks. Compute the probability that a 5-card poker hand is dealt to you that contains all diamonds two Queens. A bag contains 6 real diamonds and 5 fake diamonds. If 4 diamonds are picked from the bag at random, what is the probability that none of them are fake. exactly 2 of them are real. none of them are fake. exactly 2 of them are real.