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?

Question

Answer

0.1428

Answer

Problem Statement

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

Step 1: Calculate the number of patients who require special room facilities

One per cent of 200 patients is 2, so on average, 2 patients require special room facilities.

Step 2: Calculate the probability of 0, 1, 2, and 3 patients requiring special room facilities

Using the binomial distribution formula, we can calculate the probability of 0, 1, 2, and 3 patients requiring special room facilities:

P(0) = (0.99)^200 = 0.133

P(1) = 200C1 * (0.01)^1 * (0.99)^199 = 0.270

P(2) = 200C2 * (0.01)^2 * (0.99)^198 = 0.274

P(3) = 200C3 * (0.01)^3 * (0.99)^197 = 0.182

Step 3: Calculate the probability of more than 3 patients requiring special room facilities

The probability of more than 3 patients requiring special room facilities is the sum of the probabilities of 4, 5, 6, ..., 200 patients requiring special room facilities:

P(4 or more) = 1 - (P(0) + P(1) + P(2) + P(3))

P(4 or more) = 1 - (0.133 + 0.270 + 0.274 + 0.182)

P(4 or more) = 0.141

Step 4: Interpret the result

The probability that more than 3 patients would require special room facilities is 0.141. This means that there is a 14.1% chance that the hospital will need more than one special room on this particular morning.

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