Question
Two hospitals send doctors to a medical conference. The first hospital sends 20 doctors, and the second hospital sends 30 doctors. Only 15 doctors will be given the chance to make presentations.
What is the probability that exactly 8 of the doctors chosen to make presentations will be from the first hospital and exactly 7 of the doctors chosen to make presentations will be from the second hospital?
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Answer
The probability that exactly 8 of the doctors chosen to make presentations will be from the first hospital and exactly 7 of the doctors chosen to make presentations will be from the second hospital is approximately \(\boxed{0.1139}\).
Step 1 ：This problem involves combinations. We need to find the number of ways to choose 8 doctors from the first hospital and 7 doctors from the second hospital, and divide that by the total number of ways to choose 15 doctors from the total of 50 doctors.
Step 2 ：The number of ways to choose 8 doctors from the first hospital is \(\binom{20}{8} = 125970.0\).
Step 3 ：The number of ways to choose 7 doctors from the second hospital is \(\binom{30}{7} = 2035800.0\).
Step 4 ：The total number of ways to choose 15 doctors from the total of 50 doctors is \(\binom{50}{15} = 2250829575120.0\).
Step 5 ：The probability is calculated by dividing the product of the number of ways to choose 8 doctors from the first hospital and the number of ways to choose 7 doctors from the second hospital by the total number of ways to choose 15 doctors from the total of 50 doctors.
Step 6 ：\(\frac{125970.0 \times 2035800.0}{2250829575120.0} = 0.11393564792053514\)
Step 7 ：The probability that exactly 8 of the doctors chosen to make presentations will be from the first hospital and exactly 7 of the doctors chosen to make presentations will be from the second hospital is approximately \(\boxed{0.1139}\).
