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.
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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 \( 7.58\% \).
