Problem: 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?

Solution:

To calculate the probability that more than 3 patients would require special room facilities, we need to use the concept of the binomial distribution. The binomial distribution is used to calculate the probability of a certain number of successes (in this case, patients requiring special room facilities) in a fixed number of trials (number of patients admitted) when the probability of success (probability of requiring special room facilities) is known.

Let's break down the problem into smaller steps:

Step 1: Identify the parameters:
- Number of patients admitted daily: 200
- Probability of a patient requiring special room facilities: 1% or 0.01
- Number of patients requiring special room facilities: More than 3

Step 2: Calculate the probability of a patient requiring special room facilities:
The probability of a patient requiring special room facilities is given as 0.01 or 1%.

Step 3: Calculate the probability of more than 3 patients requiring special room facilities:
To calculate this probability, we need to find the cumulative probability of 0, 1, 2, and 3 patients requiring special room facilities and subtract it from 1.

Let's use a binomial distribution calculator or a statistical software to calculate the cumulative probability. Assuming we have access to such a tool, the probability of 0, 1, 2, and 3 patients requiring special room facilities can be calculated as follows:

- P(X = 0) = 200C0 * (0.01)^0 * (0.99)^200 = 0.1335 (approximately)
- P(X = 1) = 200C1 * (0.01)^1 * (0.99)^199 = 0.2707 (approximately)
- P(X = 2) = 200C2 * (0.01)^2 * (0.99)^198 = 0.2707 (approximately)
- P(X = 3) = 200C3 * (0.01)^3 * (0.99)^197 = 0.1805 (approximately)

The cumulative probability of more than 3 patients requiring special room facilities can be calculated as:

- P(X > 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
= 1 - (0.1335 + 0.2707 + 0.2707 + 0.1805)
= 1 - 0.8554
= 0.1446 (approximately)

Therefore, the probability that more than 3 patients would require special room facilities is approximately 0.1446 or 14.46%.

Conclusion:
The probability that more than 3 patients would require special room facilities, given that only one special room is available, is approximately 0.1446 or 14.46%.

0.1428