Answer:
(a) The probability of exactly three flaws in 150 m of cable is 0.21246
(b) The probability of at least two flaws in 100m of cable is 0.69155
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063
Step-by-step explanation:
A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by
for x = 0, 1, 2, ...
where
, the mean number of successes.
(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is ![1.2 \cdot 3 =3.6](https://tex.z-dn.net/?f=1.2%20%5Ccdot%203%20%3D3.6)
The probability of exactly three flaws in 150 m of cable is
![P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246](https://tex.z-dn.net/?f=P%28X%3D3%29%3Dp%283%3B3.6%29%3D%5Cfrac%7B3.6%5E3e%5E%7B-3.6%7D%7D%7B3%21%7D%20%5Capprox%200.21246)
(b) The probability of at least two flaws in 100m of cable is,
we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is ![1.2 \cdot 2 =2.4](https://tex.z-dn.net/?f=1.2%20%5Ccdot%202%20%3D2.4)
![P(X\geq 2)=1-P(X](https://tex.z-dn.net/?f=P%28X%5Cgeq%202%29%3D1-P%28X%3C2%29%5C%5CP%28X%5Cgeq%202%29%3D1-%28P%28X%3D0%29%2BP%28X%3D1%29%29)
![P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155](https://tex.z-dn.net/?f=P%28X%5Cgeq%202%29%3D1-p%280%3B2.4%29-p%281%3B2.4%29%5C%5C%5C%5CP%28X%5Cgeq%202%29%3D1-%5Cfrac%7B2.4%5E0e%5E%7B-2.4%7D%7D%7B0%21%7D-%5Cfrac%7B2.4%5E1e%5E%7B-2.4%7D%7D%7B1%21%7D%5C%5C%5C%5CP%28X%5Cgeq%202%29%5Capprox%200.69155)
(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is
![P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143](https://tex.z-dn.net/?f=P%28X%3D1%29%3Dp%281%3B1.2%29%3D%5Cfrac%7B1.2%5E1e%5E%7B-1.2%7D%7D%7B1%21%7D%5C%5CP%28X%3D1%29%5Capprox%200.36143)
The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is
![(0.36143)(0.36143) = 0.13063](https://tex.z-dn.net/?f=%280.36143%29%280.36143%29%20%3D%200.13063)