Answer:
Therefore, the probability is P=0.161.
Step-by-step explanation:
We know that cars arrive at a car wash randomly and independently; the probability of an arrival is the same for any two time intervals of equal length. We calculate the probability that 15 or more cars will arrive during any given hour of operation.
We know that the mean arrival rate is 10 cars per hour ⇒ λ=10.
We use the Possion formula:
[tex]\boxed{P_r(X=k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}[/tex]
we get
[tex]P(X\geq 15)=1-P(X=0)-P(X=1)-P(X=2)-...-P(X=14)\\\\P(X\geq 15)=1-\frac{10^0\cdot e^{-10}}{0!}-\frac{10^1\cdot e^{-10}}{1!}-\frac{10^2\cdot e^{-10}}{2!}-...-\frac{10^{14}\cdot e^{-10}}{14!}\\\\P(X\geq 15)=1-(0.00004+0.0004+0.002+0.007+0.01+0.03+0.06+0.09+0.1+0.1+0.12+0.11+0.09+0.07+0.05)\\\\P(X\geq 15)=1-0.839\\\\P(X\geq 15)=0.161\\[/tex]
Therefore, the probability is P=0.161.
The probability will be "0.161".
Probability seems to be a mathematical construct concerned with determining the chance of occurrence of a given event, which would be stated as a range of 1(one) as well as 0 (zero).
According to the question,
Mean arrival rate, λ = 10
By using the Possion formula,
→ [tex]P_r[/tex](X = k) = [tex]\frac{\lambda^k. e^{- \lambda}}{k!}[/tex]
Now,
The probability will be:
P(X [tex]\geq[/tex] 15) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - ... - P(X = 14)
By substituting the values,
= 1 - [tex]\frac{10^0.e^{-10}}{0!}[/tex] - [tex]\frac{10^1. e^{-10}}{1!}[/tex] - [tex]\frac{10^2.e^{-10}}{2!}[/tex] - ... - [tex]\frac{10^{14}.e^{-10}}{14!}[/tex]
= 1 - (0.00004 + 0.0004 + 0.002 + 0.007 + 0.01 + 0.03 + 0.06 + 0.09 + 0.1 + 0.1 + 0.12 + 0.11 + 0.09 + 0.07 + 0.05)
= 1 - 0.839
= 0.161
Thus the above answer is appropriate.
Find out more information about probability here:
https://brainly.com/question/24756209
