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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...
Calculating the Probability of More than 3 Patients Requiring Special Room Facilities
Probability can be calculated by determining the likelihood of an event occurring, in this case, the number of patients requiring special room facilities exceeding 3 out of the total patients admitted.
Given Information:
- Average daily admissions: 200 patients
- 1% of patients require special room facilities
- Only 1 special room is available
Step 1: Calculate the number of patients requiring special room facilities
- 1% of 200 patients = 0.01 x 200 = 2 patients requiring special room facilities on average
Step 2: Determine the probability of more than 3 patients needing special room facilities
- P(X > 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
- P(X = 0) = e^(-λ) * (λ^0) / 0! = e^(-2) * (2^0) / 0! = e^(-2) ≈ 0.1353
- P(X = 1) = e^(-2) * (2^1) / 1! = e^(-2) * 2 / 1 = 2e^(-2) ≈ 0.2707
- P(X = 2) = e^(-2) * (2^2) / 2! = e^(-2) * 4 / 2 = 2e^(-2) ≈ 0.2707
- P(X = 3) = e^(-2) * (2^3) / 3! = e^(-2) * 8 / 6 = 4e^(-2) ≈ 0.2707
- P(X > 3) = 1 - (0.1353 + 0.2707 + 0.2707 + 0.2707) ≈ 0.0526
Therefore, the probability that more than 3 patients would require special room facilities when only 1 special room is available is approximately 0.0526 or 5.26%.
Probability can be calculated by determining the likelihood of an event occurring, in this case, the number of patients requiring special room facilities exceeding 3 out of the total patients admitted.
Given Information:
- Average daily admissions: 200 patients
- 1% of patients require special room facilities
- Only 1 special room is available
Step 1: Calculate the number of patients requiring special room facilities
- 1% of 200 patients = 0.01 x 200 = 2 patients requiring special room facilities on average
Step 2: Determine the probability of more than 3 patients needing special room facilities
- P(X > 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
- P(X = 0) = e^(-λ) * (λ^0) / 0! = e^(-2) * (2^0) / 0! = e^(-2) ≈ 0.1353
- P(X = 1) = e^(-2) * (2^1) / 1! = e^(-2) * 2 / 1 = 2e^(-2) ≈ 0.2707
- P(X = 2) = e^(-2) * (2^2) / 2! = e^(-2) * 4 / 2 = 2e^(-2) ≈ 0.2707
- P(X = 3) = e^(-2) * (2^3) / 3! = e^(-2) * 8 / 6 = 4e^(-2) ≈ 0.2707
- P(X > 3) = 1 - (0.1353 + 0.2707 + 0.2707 + 0.2707) ≈ 0.0526
Therefore, the probability that more than 3 patients would require special room facilities when only 1 special room is available is approximately 0.0526 or 5.26%.
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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 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?? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 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?? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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??.
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?? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 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?? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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