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A renowned hospital usually admits 200 patients every day. One per cent patients, on an average, require special room facilities. On one particular morning, it was found that only one special room is available. What is the probability that more than 3 patients would require special room facilities?
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A renowned hospital usually admits 200 patients every day. One per cen...
Introduction
In a hospital that admits 200 patients daily, 1% typically require special room facilities. Hence, the average number of patients needing special rooms can be calculated as follows:
- Average patients needing special rooms = 1% of 200 = 2 patients.
Modeling the Situation
This scenario can be modeled using a Poisson distribution, where the average number of occurrences (λ) is equal to 2. The Poisson formula is used to find the probability of a number of events occurring in a fixed interval.
Probability of More Than 3 Patients
To calculate the probability of more than 3 patients requiring special room facilities, we need to find:
- P(X > 3) = 1 - P(X ≤ 3)
Where P(X ≤ 3) is the cumulative probability of 0, 1, 2, and 3 patients needing special rooms.
Calculating Cumulative Probability
Using the Poisson formula:
- P(X = k) = (λ^k * e^(-λ)) / k!
For k = 0, 1, 2, and 3:
- P(X = 0) = (2^0 * e^(-2)) / 0! = e^(-2)
- P(X = 1) = (2^1 * e^(-2)) / 1! = 2 * e^(-2)
- P(X = 2) = (2^2 * e^(-2)) / 2! = 2 * e^(-2)
- P(X = 3) = (2^3 * e^(-2)) / 3! = (8/6) * e^(-2)
Finally, sum these probabilities to find P(X ≤ 3) and then subtract from 1 to get P(X > 3).
Conclusion
By evaluating these probabilities, you can determine the likelihood of more than 3 patients requiring special room facilities on that particular morning. This insight is crucial for hospital management and allocation of resources.
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