Pregunta
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At the first tri-city meeting, there were 8 people from town \( A, 7 \) people from town B, and 5 people from town \( C \). If the council consists of 5 people, find the probability of 3 from town A and 2 from town B.

Ask by Chavez Boone. in the United States
Mar 20,2025

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The probability of selecting 3 people from town A and 2 people from town B in the council of 5 is \( \frac{49}{646} \).

Solución

To find the probability of 3 people from town A and 2 people from town B in the council of 5 people, we need to calculate the total number of ways to choose 3 people from town A and 2 people from town B, and then divide it by the total number of ways to choose 5 people from the total of 20 people. Let's denote: - \( A \) as the number of people from town A, - \( B \) as the number of people from town B, - \( C \) as the number of people from town C, - \( T \) as the total number of people. Given: - \( A = 8 \), - \( B = 7 \), - \( C = 5 \), - \( T = A + B + C = 8 + 7 + 5 = 20 \). We need to find the probability of 3 people from town A and 2 people from town B in the council of 5 people. The total number of ways to choose 3 people from town A is \( \binom{8}{3} \) and the total number of ways to choose 2 people from town B is \( \binom{7}{2} \). The total number of ways to choose 5 people from the total of 20 people is \( \binom{20}{5} \). Therefore, the probability of 3 people from town A and 2 people from town B in the council of 5 people is: \[ P = \frac{\binom{8}{3} \times \binom{7}{2}}{\binom{20}{5}} \] Now, let's calculate the values of the binomial coefficients and then find the probability. Calculate the value by following steps: - step0: Calculate: \( { }_{8}C_{3}\) - step1: Expand the expression: \(\frac{8!}{3!\times \left(8-3\right)!}\) - step2: Subtract the numbers: \(\frac{8!}{3!\times 5!}\) - step3: Expand the expression: \(\frac{8\times 7\times 6\times 5\times 4\times 3!}{3!\times 5!}\) - step4: Reduce the fraction: \(\frac{8\times 7\times 6\times 5\times 4}{5!}\) - step5: Multiply the terms: \(\frac{6720}{5!}\) - step6: Calculate the value: \(\frac{6720}{120}\) - step7: Reduce the fraction: \(56\) Calculate or simplify the expression \( \binom{7}{2} \). Calculate the value by following steps: - step0: Calculate: \( { }_{7}C_{2}\) - step1: Expand the expression: \(\frac{7!}{2!\times \left(7-2\right)!}\) - step2: Subtract the numbers: \(\frac{7!}{2!\times 5!}\) - step3: Expand the expression: \(\frac{7\times 6\times 5\times 4\times 3\times 2!}{2!\times 5!}\) - step4: Reduce the fraction: \(\frac{7\times 6\times 5\times 4\times 3}{5!}\) - step5: Multiply the terms: \(\frac{2520}{5!}\) - step6: Calculate the value: \(\frac{2520}{120}\) - step7: Reduce the fraction: \(21\) Calculate or simplify the expression \( \binom{20}{5} \). Calculate the value by following steps: - step0: Calculate: \( { }_{20}C_{5}\) - step1: Expand the expression: \(\frac{20!}{5!\times \left(20-5\right)!}\) - step2: Subtract the numbers: \(\frac{20!}{5!\times 15!}\) - step3: Expand the expression: \(\frac{20\times 19\times 18\times 17\times 16\times 15!}{5!\times 15!}\) - step4: Reduce the fraction: \(\frac{20\times 19\times 18\times 17\times 16}{5!}\) - step5: Multiply the terms: \(\frac{1860480}{5!}\) - step6: Calculate the value: \(\frac{1860480}{120}\) - step7: Reduce the fraction: \(15504\) Calculate or simplify the expression \( 56*21/15504 \). Calculate the value by following steps: - step0: Calculate: \(\frac{56\times 21}{15504}\) - step1: Multiply the numbers: \(\frac{1176}{15504}\) - step2: Reduce the fraction: \(\frac{49}{646}\) The probability of 3 people from town A and 2 people from town B in the council of 5 people is \( \frac{49}{646} \).

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To find the probability of selecting 3 people from town A and 2 from town B for the council of 5 people, we first need to determine the total number of ways to form the council and then the number of favorable outcomes for our specific selection. 1. **Total combinations of forming a council of 5**: The total number of people is \( 8 + 7 + 5 = 20 \). The number of ways to choose 5 out of 20 is given by the combination formula \( \binom{n}{r} \): \[ \binom{20}{5} = \frac{20!}{5!(20-5)!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 15504. \] 2. **Favorable outcomes for selecting 3 from town A and 2 from town B**: - The number of ways to choose 3 people from town A (8 people available) is: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56. \] - The number of ways to choose 2 people from town B (7 people available) is: \[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21. \] 3. **Total favorable outcomes**: The total combinations for choosing 3 from A and 2 from B is: \[ \text{Total Favorable Outcomes} = \binom{8}{3} \times \binom{7}{2} = 56 \times 21 = 1176. \] 4. **Probability calculation**: The probability of selecting 3 from town A and 2 from town B is: \[ P(3A, 2B) = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}} = \frac{1176}{15504}. \] Simplifying this fraction: \[ P(3A, 2B) = \frac{1176 \div 1176}{15504 \div 1176} = \frac{1}{13.2} \approx 0.07576. \] Thus, the probability is approximately \( 0.07576 \) or \( 7.58\% \).

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