# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total balls = **6 + 5 + 4 = 15** - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** is the ratio of the number of favorable outcomes to the total number of possible outcomes. When drawing without replacement, the number of available balls decreases after each draw. The number of ways to arrange \( n \) distinct objects is \( n! \). The number of ways to choose \( r \) objects from \( n \) is \( \binom{n}{r} \). # Solution ## Total Number of Ways to Draw 3 Balls The total number of ways to draw 3 balls (in sequence) from 15: \[ 15 \times 14 \times 13 \] ## Part 1: Probability that All Three are of Different Colors We need to select one red, one blue, and one green ball, and arrange them in any order. Number of ways to choose one red: **6** Number of ways to choose one blue: **5** Number of ways to choose one green: **4** Number of ways to arrange the 3 balls: **3! = 6** So, total favorable ways: \[ 6 \times 5 \times 4 \times 6 = 720 \] Probability: \[ P(\text{3 different colors}) = \frac{720}{15 \times 14 \times 13} \] Calculate denominator: \[ 15 \times 14 \times 13 = 273 \] \[ P(\text{3 different colors}) = \frac{720}{273} = \frac{72}{273} = \frac{24}{91} \] ## Part 2: Probability that Exactly Two are of the Same Color This event can happen in the following ways: - Two balls of the same color, one ball of a different color. Let's consider each color pair: ### Case 1: Two red and one non-red Ways to pick 2 reds: \( \binom{6}{2} = 15 \) Ways to pick 1 non-red (either blue or green): \( 5 + 4 = 9 \) Number of arrangements: \( \frac{3!}{2!} = 3 \) (since 2 are of the same color, 1 different) Total ways for this case: \[ 15 \times 9 \times 3 = 405 \] ### Case 2: Two blue and one non-blue Ways to pick 2 blue: \( \binom{5}{2} = 10 \) Ways to pick 1 non-blue (red or green): \( 6 + 4 = 10 \) Arrangements: 3 Total ways: \[ 10 \times 10 \times 3 = 300 \] ### Case 3: Two green and one non-green Ways to pick 2 green: \( \binom{4}{2} = 6 \) Ways to pick 1 non-green (red or blue): \( 6 + 5 = 11 \) Arrangements: 3 Total ways: \[ 6 \times 11 \times 3 = 198 \] Add all cases: \[ 405 + 300 + 198 = 903 \] Probability: \[ P(\text{exactly 2 same color}) = \frac{903}{273} = \frac{301}{910} \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{301}{910}} \]

# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total balls = **6 + 5 + 4 = 15** - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** is the ratio of the number of favorable outcomes to the total number of possible outcomes. When drawing without replacement, the number of available balls decreases after each draw. The number of ways to arrange \( n \) distinct objects is \( n! \). The number of ways to choose \( r \) objects from \( n \) is \( \binom{n}{r} \). # Solution ## Total Number of Ways to Draw 3 Balls The total number of ways to draw 3 balls (in sequence) from 15: \[ 15 \times 14 \times 13 = 2730 \] ## Part 1: Probability that All Three are of Different Colors We need to select one red, one blue, and one green ball, and arrange them in any order. Number of ways to choose one red: **6** Number of ways to choose one blue: **5** Number of ways to choose one green: **4** Number of ways to arrange the 3 balls: **3! = 6** Total favorable ways: \[ 6 \times 5 \times 4 \times 6 = 720 \] Probability: \[ P(\text{3 different colors}) = \frac{720}{2730} \] Calculating the fraction: \[ P(\text{3 different colors}) = \frac{720}{2730} \approx 0.2637 \] ## Part 2: Probability that Exactly Two are of the Same Color This event can happen in the following ways: - Two balls of the same color, one ball of a different color. Let's consider each color pair: ### Case 1: Two red and one non-red Ways to pick 2 reds: \( \binom{6}{2} = 15 \) Ways to pick 1 non-red (either blue or green): \( 5 + 4 = 9 \) Number of arrangements: \( \frac{3!}{2!} = 3 \) (since 2 are of the same color, 1 different) Total ways for this case: \[ 15 \times 9 \times 3 = 405 \] ### Case 2: Two blue and one non-blue Ways to pick 2 blue: \( \binom{5}{2} = 10 \) Ways to pick 1 non-blue (red or green): \( 6 + 4 = 10 \) Arrangements: 3 Total ways: \[ 10 \times 10 \times 3 = 300 \] ### Case 3: Two green and one non-green Ways to pick 2 green: \( \binom{4}{2} = 6 \) Ways to pick 1 non-green (red or blue): \( 6 + 5 = 11 \) Arrangements: 3 Total ways: \[ 6 \times 11 \times 3 = 198 \] Add all cases: \[ 405 + 300 + 198 = 903 \] Probability: \[ P(\text{exactly 2 same color}) = \frac{903}{2730} \] Calculating the fraction: \[ P(\text{exactly 2 same color}) = \frac{903}{2730} \approx 0.3307 \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{301}{910}} \approx 0.3307 \]

# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total balls = **6 + 5 + 4 = 15** - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** is the ratio of the number of favorable outcomes to the total number of possible outcomes. When drawing without replacement, the number of available balls decreases after each draw. The number of ways to choose \( r \) objects from \( n \) is given by the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total number of ways to draw 3 balls from 15 is: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three are of Different Colors To have all three balls of different colors, we need to choose one ball from each color. - Choose 1 red from 6: \( \binom{6}{1} = 6 \) - Choose 1 blue from 5: \( \binom{5}{1} = 5 \) - Choose 1 green from 4: \( \binom{4}{1} = 4 \) The total ways to choose one of each color: \[ 6 \times 5 \times 4 = 120 \] Since the order matters, we multiply by the number of arrangements of 3 distinct balls, which is \( 3! = 6 \): \[ 120 \times 6 = 720 \] Probability: \[ P(\text{3 different colors}) = \frac{720}{455} = \frac{144}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two are of the Same Color This event can occur in three scenarios: 1. Two red and one other color. 2. Two blue and one other color. 3. Two green and one other color. ### Case 1: Two Red and One Non-Red - Choose 2 reds from 6: \( \binom{6}{2} = 15 \) - Choose 1 non-red (either blue or green): \( \binom{9}{1} = 9 \) (5 blue + 4 green) - Arrangements: \( \frac{3!}{2!} = 3 \) Total ways for this case: \[ 15 \times 9 \times 3 = 405 \] ### Case 2: Two Blue and One Non-Blue - Choose 2 blues from 5: \( \binom{5}{2} = 10 \) - Choose 1 non-blue (either red or green): \( \binom{10}{1} = 10 \) (6 red + 4 green) - Arrangements: 3 Total ways: \[ 10 \times 10 \times 3 = 300 \] ### Case 3: Two Green and One Non-Green - Choose 2 greens from 4: \( \binom{4}{2} = 6 \) - Choose 1 non-green (either red or blue): \( \binom{11}{1} = 11 \) (6 red + 5 blue) - Arrangements: 3 Total ways: \[ 6 \times 11 \times 3 = 198 \] Add all cases: \[ 405 + 300 + 198 = 903 \] Probability: \[ P(\text{exactly 2 same color}) = \frac{903}{455} \approx 0.1980 \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{903}{455}} \approx 0.1980 \]

# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total balls = **6 + 5 + 4 = 15** - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** is the ratio of the number of favorable outcomes to the total number of possible outcomes. When drawing without replacement, the number of available balls decreases after each draw. **Combination** is used to calculate the number of ways to choose \( r \) objects from \( n \) without regard to the order, represented as: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total number of ways to draw 3 balls from 15 is: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three are of Different Colors To have all three balls of different colors, we need to choose one ball from each color. - Choose 1 red from 6: \( \binom{6}{1} = 6 \) - Choose 1 blue from 5: \( \binom{5}{1} = 5 \) - Choose 1 green from 4: \( \binom{4}{1} = 4 \) The total ways to choose one of each color: \[ 6 \times 5 \times 4 = 120 \] Since the order does not matter, we do not need to multiply by arrangements. Probability: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two are of the Same Color This event can occur in three scenarios: 1. Two red and one other color. 2. Two blue and one other color. 3. Two green and one other color. ### Case 1: Two Red and One Non-Red - Choose 2 reds from 6: \( \binom{6}{2} = 15 \) - Choose 1 non-red (either blue or green): \( \binom{9}{1} = 9 \) (5 blue + 4 green) Total ways for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - Choose 2 blues from 5: \( \binom{5}{2} = 10 \) - Choose 1 non-blue (either red or green): \( \binom{10}{1} = 10 \) (6 red + 4 green) Total ways: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - Choose 2 greens from 4: \( \binom{4}{2} = 6 \) - Choose 1 non-green (either red or blue): \( \binom{11}{1} = 11 \) (6 red + 5 blue) Total ways: \[ 6 \times 11 = 66 \] Add all cases: \[ 135 + 100 + 66 = 301 \] Probability: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total number of balls = **15** (6 + 5 + 4) - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. **Combination** is a selection of items without considering the order, calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total number of ways to select 3 balls from 15 is given by: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three are of Different Colors To find the probability that all three drawn balls are of different colors, we will select one ball from each color. - Choose 1 red from 6: \( \binom{6}{1} = 6 \) - Choose 1 blue from 5: \( \binom{5}{1} = 5 \) - Choose 1 green from 4: \( \binom{4}{1} = 4 \) Calculating the total combinations: \[ 6 \times 5 \times 4 = 120 \] The probability is then: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two are of the Same Color This scenario can occur in three distinct cases: 1. Two red and one other color. 2. Two blue and one other color. 3. Two green and one other color. ### Case 1: Two Red and One Non-Red - Choose 2 reds from 6: \[ \binom{6}{2} = 15 \] - Choose 1 non-red (blue or green): \[ \binom{9}{1} = 9 \quad (\text{5 blue + 4 green}) \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - Choose 2 blues from 5: \[ \binom{5}{2} = 10 \] - Choose 1 non-blue (red or green): \[ \binom{10}{1} = 10 \quad (\text{6 red + 4 green}) \] Total combinations: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - Choose 2 greens from 4: \[ \binom{4}{2} = 6 \] - Choose 1 non-green (red or blue): \[ \binom{11}{1} = 11 \quad (\text{6 red + 5 blue}) \] Total combinations: \[ 6 \times 11 = 66 \] Adding the total combinations from all cases gives: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total number of balls = **15** (6 + 5 + 4) - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** is the ratio of favorable outcomes to the total possible outcomes. **Combination** is the selection of items where the order does not matter, calculated using: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total ways to choose 3 balls from 15 is given by: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three are of Different Colors To calculate the probability that all three balls drawn are of different colors, we need to choose one ball from each color. - Choose 1 red from 6: \( \binom{6}{1} = 6 \) - Choose 1 blue from 5: \( \binom{5}{1} = 5 \) - Choose 1 green from 4: \( \binom{4}{1} = 4 \) The total combinations for selecting one of each color is: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two are of the Same Color This situation can occur in three distinct cases: 1. Two red and one of another color. 2. Two blue and one of another color. 3. Two green and one of another color. ### Case 1: Two Red and One Non-Red - Choose 2 reds from 6: \[ \binom{6}{2} = 15 \] - Choose 1 non-red (blue or green): \[ \binom{9}{1} = 9 \quad (\text{5 blue + 4 green}) \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - Choose 2 blues from 5: \[ \binom{5}{2} = 10 \] - Choose 1 non-blue (red or green): \[ \binom{10}{1} = 10 \quad (\text{6 red + 4 green}) \] Total combinations: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - Choose 2 greens from 4: \[ \binom{4}{2} = 6 \] - Choose 1 non-green (red or blue): \[ \binom{11}{1} = 11 \quad (\text{6 red + 5 blue}) \] Total combinations: \[ 6 \times 11 = 66 \] Adding the combinations from all cases gives: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total number of balls = **15** (6 + 5 + 4) - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** is defined as the ratio of the number of favorable outcomes to the total number of possible outcomes. **Combination** is a selection of items where the order does not matter, calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total ways to select 3 balls from 15 can be calculated using: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three are of Different Colors To determine the probability that all three balls drawn are of different colors, we select one ball from each color. - Choose 1 red from 6: \( \binom{6}{1} = 6 \) - Choose 1 blue from 5: \( \binom{5}{1} = 5 \) - Choose 1 green from 4: \( \binom{4}{1} = 4 \) The total combinations for selecting one ball of each color is: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two are of the Same Color This scenario can occur in three distinct cases: 1. Two red and one of another color. 2. Two blue and one of another color. 3. Two green and one of another color. ### Case 1: Two Red and One Non-Red - Choose 2 reds from 6: \[ \binom{6}{2} = 15 \] - Choose 1 non-red (either blue or green): \[ \binom{9}{1} = 9 \quad (\text{5 blue + 4 green}) \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - Choose 2 blues from 5: \[ \binom{5}{2} = 10 \] - Choose 1 non-blue (either red or green): \[ \binom{10}{1} = 10 \quad (\text{6 red + 4 green}) \] Total combinations: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - Choose 2 greens from 4: \[ \binom{4}{2} = 6 \] - Choose 1 non-green (either red or blue): \[ \binom{11}{1} = 11 \quad (\text{6 red + 5 blue}) \] Total combinations: \[ 6 \times 11 = 66 \] Adding the combinations from all cases gives: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - Number of red balls = **6** - Number of blue balls = **5** - Number of green balls = **4** - Total number of balls = **15** (6 + 5 + 4) - Three balls are drawn **one by one without replacement** # Definitions and Concepts **Probability** represents the likelihood of an event, expressed as the ratio of favorable outcomes to the total possible outcomes. **Combinations** are used to determine the number of ways to select items without regard to the order, calculated by: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total combinations of selecting 3 balls from 15 can be calculated as: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three are of Different Colors To find the probability that all three balls drawn are of different colors, we need to select one ball from each color. - Choose 1 red from 6: \( \binom{6}{1} = 6 \) - Choose 1 blue from 5: \( \binom{5}{1} = 5 \) - Choose 1 green from 4: \( \binom{4}{1} = 4 \) The total combinations for selecting one ball of each color is: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two are of the Same Color This situation can occur in three cases: 1. Two red and one different color. 2. Two blue and one different color. 3. Two green and one different color. ### Case 1: Two Red and One Non-Red - Choose 2 reds from 6: \[ \binom{6}{2} = 15 \] - Choose 1 non-red (either blue or green): \[ \binom{9}{1} = 9 \quad (\text{5 blue + 4 green}) \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - Choose 2 blues from 5: \[ \binom{5}{2} = 10 \] - Choose 1 non-blue (either red or green): \[ \binom{10}{1} = 10 \quad (\text{6 red + 4 green}) \] Total combinations: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - Choose 2 greens from 4: \[ \binom{4}{2} = 6 \] - Choose 1 non-green (either red or blue): \[ \binom{11}{1} = 11 \quad (\text{6 red + 5 blue}) \] Total combinations: \[ 6 \times 11 = 66 \] The total combinations from all cases: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - **Number of Red Balls:** 6 - **Number of Blue Balls:** 5 - **Number of Green Balls:** 4 - **Total Number of Balls:** 15 (6 + 5 + 4) - **Drawing Method:** Three balls drawn one by one without replacement # Definitions and Concepts **Probability** quantifies the likelihood of an event occurring, expressed as the ratio of favorable outcomes to the total number of possible outcomes. **Combination** refers to selecting items without considering the order, calculated using: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls To determine the total combinations for selecting 3 balls from 15: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three Balls are of Different Colors To achieve different colors for all three drawn balls, one ball must be selected from each color category. - **Choosing 1 Red:** \[ \binom{6}{1} = 6 \] - **Choosing 1 Blue:** \[ \binom{5}{1} = 5 \] - **Choosing 1 Green:** \[ \binom{4}{1} = 4 \] The total combinations of selecting one ball from each color: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability of drawing three balls of different colors is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two Balls are of the Same Color This scenario can manifest in three distinct ways: 1. Two reds and one of another color. 2. Two blues and one of another color. 3. Two greens and one of another color. ### Case 1: Two Red and One Non-Red - **Choosing 2 Reds:** \[ \binom{6}{2} = 15 \] - **Choosing 1 Non-Red (Blue or Green):** \[ \binom{9}{1} = 9 \quad (\text{5 blue + 4 green}) \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - **Choosing 2 Blues:** \[ \binom{5}{2} = 10 \] - **Choosing 1 Non-Blue (Red or Green):** \[ \binom{10}{1} = 10 \quad (\text{6 red + 4 green}) \] Total combinations for this case: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - **Choosing 2 Greens:** \[ \binom{4}{2} = 6 \] - **Choosing 1 Non-Green (Red or Blue):** \[ \binom{11}{1} = 11 \quad (\text{6 red + 5 blue}) \] Total combinations for this case: \[ 6 \times 11 = 66 \] Adding the combinations from all cases gives: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three balls are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two balls are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - **Number of Red Balls:** 6 - **Number of Blue Balls:** 5 - **Number of Green Balls:** 4 - **Total Number of Balls:** 15 (6 + 5 + 4) - **Drawing Method:** Three balls drawn sequentially without replacement # Definitions and Concepts **Probability** measures the chance of an event occurring, expressed as the ratio of favorable outcomes to the total possible outcomes. **Combination** refers to selecting items without considering the order, calculated with: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total number of ways to choose 3 balls from 15 can be calculated as follows: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three Balls are of Different Colors To ensure that all three drawn balls are of different colors, one ball must be selected from each color. - **Choosing 1 Red:** \[ \binom{6}{1} = 6 \] - **Choosing 1 Blue:** \[ \binom{5}{1} = 5 \] - **Choosing 1 Green:** \[ \binom{4}{1} = 4 \] The total combinations for selecting one ball from each color is: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability of drawing three balls of different colors is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two Balls are of the Same Color This scenario can unfold in three distinct cases: 1. Two reds and one of another color. 2. Two blues and one of another color. 3. Two greens and one of another color. ### Case 1: Two Red and One Non-Red - **Choosing 2 Reds:** \[ \binom{6}{2} = 15 \] - **Choosing 1 Non-Red (Blue or Green):** \[ \binom{9}{1} = 9 \quad (\text{5 blue + 4 green}) \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - **Choosing 2 Blues:** \[ \binom{5}{2} = 10 \] - **Choosing 1 Non-Blue (Red or Green):** \[ \binom{10}{1} = 10 \quad (\text{6 red + 4 green}) \] Total combinations for this case: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - **Choosing 2 Greens:** \[ \binom{4}{2} = 6 \] - **Choosing 1 Non-Green (Red or Blue):** \[ \binom{11}{1} = 11 \quad (\text{6 red + 5 blue}) \] Total combinations for this case: \[ 6 \times 11 = 66 \] Adding the combinations from all cases results in: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three balls are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two balls are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - **Number of Red Balls:** 6 - **Number of Blue Balls:** 5 - **Number of Green Balls:** 4 - **Total Number of Balls:** 15 (6 + 5 + 4) - **Drawing Method:** Three balls drawn sequentially without replacement # Definitions and Concepts **Probability** quantifies the likelihood of an event occurring, expressed as the ratio of favorable outcomes to the total number of possible outcomes. **Combination** refers to selecting items without considering the order, calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] # Solution ## Total Number of Ways to Draw 3 Balls The total combinations for selecting 3 balls from 15 is calculated as: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three Balls are of Different Colors To achieve different colors for all three drawn balls, one ball must be selected from each color category. - **Choose 1 Red:** \[ \binom{6}{1} = 6 \] - **Choose 1 Blue:** \[ \binom{5}{1} = 5 \] - **Choose 1 Green:** \[ \binom{4}{1} = 4 \] Calculating the total combinations for selecting one of each color: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two Balls are of the Same Color This scenario can occur in three distinct ways: 1. Two reds and one of another color. 2. Two blues and one of another color. 3. Two greens and one of another color. ### Case 1: Two Red and One Non-Red - **Choose 2 Reds:** \[ \binom{6}{2} = 15 \] - **Choose 1 Non-Red (Blue or Green):** \[ \binom{9}{1} = 9 \quad (\text{5 blue + 4 green}) \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue and One Non-Blue - **Choose 2 Blues:** \[ \binom{5}{2} = 10 \] - **Choose 1 Non-Blue (Red or Green):** \[ \binom{10}{1} = 10 \quad (\text{6 red + 4 green}) \] Total combinations for this case: \[ 10 \times 10 = 100 \] ### Case 3: Two Green and One Non-Green - **Choose 2 Greens:** \[ \binom{4}{2} = 6 \] - **Choose 1 Non-Green (Red or Blue):** \[ \binom{11}{1} = 11 \quad (\text{6 red + 5 blue}) \] Total combinations for this case: \[ 6 \times 11 = 66 \] Adding the combinations from all cases gives: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three balls are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two balls are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - **Number of Red Balls:** 6 - **Number of Blue Balls:** 5 - **Number of Green Balls:** 4 - **Total Number of Balls:** 15 (6 + 5 + 4) - **Drawing Method:** Three balls drawn sequentially without replacement # Definitions and Concepts **Probability** measures the likelihood of a specific event occurring compared to the total possible outcomes. It is expressed as: \[ P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} \] **Combination** is a mathematical way to select items without regard to the order of selection. It is calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] This formula is useful when dealing with problems where order does not matter, such as selecting balls of different colors from a collection. # Solution ## Total Number of Ways to Draw 3 Balls To find the total ways to select 3 balls from the 15 available, we use the combination formula: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three Balls are of Different Colors To determine the probability that all three drawn balls are of different colors, we need one ball from each color category. The calculation is as follows: 1. **Choose 1 Red Ball:** The number of ways to choose 1 red ball from 6: \[ \binom{6}{1} = 6 \] 2. **Choose 1 Blue Ball:** The number of ways to choose 1 blue ball from 5: \[ \binom{5}{1} = 5 \] 3. **Choose 1 Green Ball:** The number of ways to choose 1 green ball from 4: \[ \binom{4}{1} = 4 \] The total combinations for selecting one ball of each color is: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability of selecting three balls of different colors is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two Balls are of the Same Color This scenario can occur in three distinct ways, depending on which color is represented twice: 1. **Two Reds and One Other Color** 2. **Two Blues and One Other Color** 3. **Two Greens and One Other Color** ### Case 1: Two Red Balls and One Non-Red - **Choose 2 Red Balls:** The number of ways to choose 2 red balls from 6: \[ \binom{6}{2} = 15 \] - **Choose 1 Non-Red Ball (Blue or Green):** The number of non-red balls is 9 (5 blue + 4 green): \[ \binom{9}{1} = 9 \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue Balls and One Non-Blue - **Choose 2 Blue Balls:** The number of ways to choose 2 blue balls from 5: \[ \binom{5}{2} = 10 \] - **Choose 1 Non-Blue Ball (Red or Green):** The number of non-blue balls is 10 (6 red + 4 green): \[ \binom{10}{1} = 10 \] Total combinations for this case: \[ 10 \times 10 = 100 \] ### Case 3: Two Green Balls and One Non-Green - **Choose 2 Green Balls:** The number of ways to choose 2 green balls from 4: \[ \binom{4}{2} = 6 \] - **Choose 1 Non-Green Ball (Red or Blue):** The number of non-green balls is 11 (6 red + 5 blue): \[ \binom{11}{1} = 11 \] Total combinations for this case: \[ 6 \times 11 = 66 \] ### Total Combinations for Exactly Two Balls of the Same Color Adding the combinations from all three cases gives: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three balls are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two balls are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]

# Given Information - **Number of Red Balls:** 6 - **Number of Blue Balls:** 5 - **Number of Green Balls:** 4 - **Total Number of Balls:** 15 (6 + 5 + 4) - **Drawing Method:** Three balls drawn sequentially without replacement # Definitions and Concepts **Probability** is the measure of the likelihood of an event occurring, expressed as the ratio of the number of favorable outcomes to the total number of possible outcomes. **Combination** refers to the selection of items without regard to the order of selection, calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] This formula is particularly useful for problems involving selections from groups where the order does not matter, such as drawing colored balls from a collection. # Solution ## Total Number of Ways to Draw 3 Balls To find the total ways to select 3 balls from the 15 available, we can apply the combination formula: \[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455 \] ## Part 1: Probability that All Three Balls are of Different Colors To determine the probability that all three drawn balls are of different colors, we need to select one ball from each color category. The calculation is as follows: 1. **Choose 1 Red Ball:** The number of ways to choose 1 red ball from 6: \[ \binom{6}{1} = 6 \] 2. **Choose 1 Blue Ball:** The number of ways to choose 1 blue ball from 5: \[ \binom{5}{1} = 5 \] 3. **Choose 1 Green Ball:** The number of ways to choose 1 green ball from 4: \[ \binom{4}{1} = 4 \] The total combinations for selecting one ball of each color is: \[ 6 \times 5 \times 4 = 120 \] Thus, the probability of selecting three balls of different colors is: \[ P(\text{3 different colors}) = \frac{120}{455} = \frac{24}{91} \approx 0.2637 \] ## Part 2: Probability that Exactly Two Balls are of the Same Color This scenario can occur in three distinct ways, depending on which color is represented twice: 1. **Two Reds and One Other Color** 2. **Two Blues and One Other Color** 3. **Two Greens and One Other Color** ### Case 1: Two Red Balls and One Non-Red - **Choose 2 Red Balls:** The number of ways to choose 2 red balls from 6: \[ \binom{6}{2} = 15 \] - **Choose 1 Non-Red Ball (Blue or Green):** The number of non-red balls is 9 (5 blue + 4 green): \[ \binom{9}{1} = 9 \] Total combinations for this case: \[ 15 \times 9 = 135 \] ### Case 2: Two Blue Balls and One Non-Blue - **Choose 2 Blue Balls:** The number of ways to choose 2 blue balls from 5: \[ \binom{5}{2} = 10 \] - **Choose 1 Non-Blue Ball (Red or Green):** The number of non-blue balls is 10 (6 red + 4 green): \[ \binom{10}{1} = 10 \] Total combinations for this case: \[ 10 \times 10 = 100 \] ### Case 3: Two Green Balls and One Non-Green - **Choose 2 Green Balls:** The number of ways to choose 2 green balls from 4: \[ \binom{4}{2} = 6 \] - **Choose 1 Non-Green Ball (Red or Blue):** The number of non-green balls is 11 (6 red + 5 blue): \[ \binom{11}{1} = 11 \] Total combinations for this case: \[ 6 \times 11 = 66 \] ### Total Combinations for Exactly Two Balls of the Same Color Adding the combinations from all three cases gives: \[ 135 + 100 + 66 = 301 \] The probability of drawing exactly two balls of the same color is: \[ P(\text{exactly 2 same color}) = \frac{301}{455} \approx 0.6615 \] # Final Solutions - **Probability that all three balls are of different colors:** \[ \boxed{\frac{24}{91}} \approx 0.2637 \] - **Probability that exactly two balls are of the same color:** \[ \boxed{\frac{301}{455}} \approx 0.6615 \]