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. Â 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)Â 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
