Questions: 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?

Solution Steps

First, we calculate the number of ways to choose 8 doctors from the first hospital and 7 doctors from the second hospital using combinations:

\[ \text{ways\_first\_hospital} = \binom{20}{8} = 125970 \]

\[ \text{ways\_second\_hospital} = \binom{30}{7} = 2035800 \]

Next, we calculate the total number of ways to choose 15 doctors from the combined pool of 50 doctors (20 from the first hospital and 30 from the second hospital):

\[ \text{total\_ways} = \binom{50}{15} = 2250829575120 \]

The probability that exactly 8 of the doctors chosen to make presentations will be from the first hospital and exactly 7 will be from the second hospital is given by:

\[ \text{probability} = \frac{\text{ways\_first\_hospital} \times \text{ways\_second\_hospital}}{\text{total\_ways}} \]

Substituting the values:

\[ \text{probability} = \frac{125970 \times 2035800}{2250829575120} \approx 0.1139 \]

Final Answer