Answer:
a)P= 0.139
b)P(7)= 0.0352
Explanation:
Given that
Number of messages per day ,n= 2500
The probability that a message will fail,p=0.005
Here n is much much larger than the P that is why we will use Poisson distribution
In Poisson distribution
Mean ,λ= n p
λ= 2500 x 0.005
λ=12.5
a)
The probability that fewer than 2485 messages reach the base station =P
[tex]P(x)=\dfrac{\lambda ^x.e^{-\lambda }}{x!}[/tex]
P= 1 - (P(0)+P(1)+P(2) ----------+P(15))
[tex]P(0)=\dfrac{12.5 ^0.e^{-1.25 }}{0!}[/tex]
[tex]P(1)=\dfrac{12.5 ^1.e^{-1.25 }}{1!}[/tex]
[tex]P(2)=\dfrac{12.5 ^2.e^{-1.25 }}{2!}[/tex]
[tex]P(3)=\dfrac{12.5 ^3.e^{-1.25 }}{3!}[/tex]
[tex]P=1-\left( \dfrac{12.5 ^0.e^{-1.25 }}{0!}+\dfrac{12.5 ^1.e^{-1.25 }}{1!}+\dfrac{12.5 ^2.e^{-1.25 }}{2!} +\dfrac{12.5 ^3.e^{-1.25 }}{3!}-----+\dfrac{12.5 ^{15}.e^{-1.25 }}{15!} \right)[/tex]
By solving this
P= 0.139
b)
The probability that 7 of the messages fail to reach the base station=P(7)
[tex]P(7)=\dfrac{12.5 ^7.e^{-1.25 }}{7!}[/tex]
P(7)= 0.0352
