3医生4病人场景下每位医生均被呼叫的概率求解及模型咨询

Great question! Let's walk through validating your probability model and computing the desired probability step by step.

First: Your Probability Model Is Correct (Let's Fill in the Gaps) #

You started defining the key sets, so let's formalize the rest to make it complete:

• Doctor set: ( D = {d_1, d_2, d_3} ) (3 total doctors)
• Patient set: ( M = {m_1, m_2, m_3, m_4} ) (4 total patients)
• Sample space ( \Omega = D^M ): This represents all possible ways patients can call doctors—each element is a tuple where every entry corresponds to which doctor a specific patient called. Since each patient has 3 independent choices, the total number of outcomes in the sample space is ( |\Omega| = 3^4 = 81 ).
• Probability measure ( P ): Assuming every patient picks a doctor uniformly at random (all outcomes are equally likely), the probability of any event ( A \subseteq \Omega ) is calculated as:
  [
  P(A) = \frac{\text{Number of favorable outcomes in } A}{|\Omega|}
  ]

Next: Calculate the Probability All Doctors Are Called #

We care about the event ( A ): Every doctor is called by at least one patient. To find how many outcomes fall into this event, we can use two straightforward methods:

Method 1: Inclusion-Exclusion Principle

This approach works by subtracting the "bad" outcomes (where at least one doctor is never called) from the total number of outcomes:

• Let ( A_i ) = the set of assignments where doctor ( d_i ) is never called.
• We need to compute ( |\Omega| - |A_1 \cup A_2 \cup A_3| ).

Using the inclusion-exclusion principle:
[
|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|
]

• ( |A_i| = 2^4 = 16 ): If one doctor is ignored, each patient picks from the remaining 2 doctors. There are 3 such sets, so their sum is ( 3 \times 16 = 48 ).
• ( |A_i \cap A_j| = 1^4 = 1 ): If two doctors are ignored, all patients must call the single remaining doctor. There are 3 such pairs, so their sum is ( 3 \times 1 = 3 ).
• ( |A_1 \cap A_2 \cap A_3| = 0 ): This is impossible—every patient has to call some doctor, so there are no outcomes where all three doctors are ignored.

Plugging these values in:
[
|A_1 \cup A_2 \cup A_3| = 48 - 3 + 0 = 45
]
So the number of favorable outcomes is ( |A| = 81 - 45 = 36 ).

Method 2: Stirling Numbers of the Second Kind

We can also count the number of surjective functions (assignments where every doctor gets at least one patient) using Stirling numbers. The formula for the number of surjective functions from a set of size ( n ) to a set of size ( k ) is:
[
\text{Number of surjective functions} = k! \times S(n, k)
]
Where ( S(n, k) ) is the Stirling number of the second kind (the number of ways to partition ( n ) elements into ( k ) non-empty subsets). For ( n=4 ) (patients) and ( k=3 ) (doctors), ( S(4, 3) = 6 )—this comes from choosing 2 patients to group together (the other two are singletons), which is ( \binom{4}{2} = 6 ).

Calculating:
[
|A| = 3! \times 6 = 6 \times 6 = 36
]

Final Probability #

Now we can compute the probability that every doctor is called:
[
P(A) = \frac{36}{81} = \frac{4}{9} \approx 0.444
]

内容的提问来源于stack exchange，提问作者Vinícius Lopes Simões