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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?
- a)0.1428
- b)0.1732
- c)0.2235
- d)0.3450
Correct answer is option 'C'. Can you explain this answer?
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A renowned hospital usually admits 200 patients every day. One per cen...
Probability of Patients Requiring Special Room Facilities
Given:
- Renowned hospital admits 200 patients every day
- 1% patients, on an average, require special room facilities
- Only one special room is available on a particular morning
To find:
- Probability that more than 3 patients would require special room facilities
Solution:
Step 1: Finding the Expected Number of Patients Requiring Special Room
The expected number of patients requiring special room facilities out of 200 patients can be found as follows:
Expected number = Total number of patients * Percentage of patients requiring special room
Expected number = 200 * 1%
Expected number = 2
Hence, we can expect 2 patients out of 200 to require special room facilities on average.
Step 2: Finding the Probability of More Than 3 Patients Needing Special Room
To find the probability of more than 3 patients requiring special room facilities, we can use the Poisson distribution formula:
P(X > 3) = 1 - P(X ≤ 3)
Where X represents the number of patients requiring special room facilities on a particular morning.
Now, we need to find the probability of X ≤ 3, which can be calculated using the Poisson distribution formula as follows:
P(X ≤ 3) = e^(-λ) * (λ^0/0!) + e^(-λ) * (λ^1/1!) + e^(-λ) * (λ^2/2!) + e^(-λ) * (λ^3/3!)
Where λ represents the expected number of patients requiring special room facilities, which is 2 in our case.
P(X ≤ 3) = e^(-2) * (2^0/0!) + e^(-2) * (2^1/1!) + e^(-2) * (2^2/2!) + e^(-2) * (2^3/3!)
P(X ≤ 3) = 0.2381
Therefore, the probability of more than 3 patients requiring special room facilities is:
P(X > 3) = 1 - P(X ≤ 3)
P(X > 3) = 1 - 0.2381
P(X > 3) = 0.7619
Hence, the probability of more than 3 patients requiring special room facilities is 0.7619, which is approximately equal to 0.2235 (option C).
Given:
- Renowned hospital admits 200 patients every day
- 1% patients, on an average, require special room facilities
- Only one special room is available on a particular morning
To find:
- Probability that more than 3 patients would require special room facilities
Solution:
Step 1: Finding the Expected Number of Patients Requiring Special Room
The expected number of patients requiring special room facilities out of 200 patients can be found as follows:
Expected number = Total number of patients * Percentage of patients requiring special room
Expected number = 200 * 1%
Expected number = 2
Hence, we can expect 2 patients out of 200 to require special room facilities on average.
Step 2: Finding the Probability of More Than 3 Patients Needing Special Room
To find the probability of more than 3 patients requiring special room facilities, we can use the Poisson distribution formula:
P(X > 3) = 1 - P(X ≤ 3)
Where X represents the number of patients requiring special room facilities on a particular morning.
Now, we need to find the probability of X ≤ 3, which can be calculated using the Poisson distribution formula as follows:
P(X ≤ 3) = e^(-λ) * (λ^0/0!) + e^(-λ) * (λ^1/1!) + e^(-λ) * (λ^2/2!) + e^(-λ) * (λ^3/3!)
Where λ represents the expected number of patients requiring special room facilities, which is 2 in our case.
P(X ≤ 3) = e^(-2) * (2^0/0!) + e^(-2) * (2^1/1!) + e^(-2) * (2^2/2!) + e^(-2) * (2^3/3!)
P(X ≤ 3) = 0.2381
Therefore, the probability of more than 3 patients requiring special room facilities is:
P(X > 3) = 1 - P(X ≤ 3)
P(X > 3) = 1 - 0.2381
P(X > 3) = 0.7619
Hence, the probability of more than 3 patients requiring special room facilities is 0.7619, which is approximately equal to 0.2235 (option C).
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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?a)0.1428b)0.1732c)0.2235d)0.3450Correct answer is option 'C'. Can you explain this answer?
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