It as been found that on an average 4 patients visit a particular doctor during 1 hour. what is the probability that during a particular hour. 1) doctor is free. 2) more than 3 patients visit the doctor?
Answers
Given : on an average 4 patients visit a particular doctor during 1 hour
To find : probability that during a particular hour. 1) doctor is free. 2) more than 3 patients visit the doctor
Solution:
on an average 4 patients visit a particular doctor during 1 hour
λ = 4
P(x) =( λˣ e^(-λ) ) / x!
1) doctor is free. => 0 patient visited Doctor
P(0) = 4⁰ e⁻⁴ / 0! = 0.0183
probability that during a particular hour doctor is free. = 0.0183
more than 3 patients visit the doctor
= 1 - P(0) - P(1) - P(2) - P(3)
= 1 - 4⁰ e⁻⁴ / 0! - 4¹e⁻⁴ /1! - 4²e⁻⁴ /2! - 4³e⁻⁴/3!
= 1 - ( 0.0183 + 0.0733 + 0.1465 + 0.1954)
= 1 - (0.4335)
= 0.5665
probability that during a particular hour more than 3 patients visit the doctor
= 0.5665
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