Answer:
0.715
Explanation:
To find the probability of at least 5 power outages in any three-month period when power outages occur at a rate of two per month, you can use the Poisson distribution. The Poisson distribution is often used to model the number of events happening in a fixed interval of time when these events occur with a known average rate.
In this case, the average rate of power outages per month is 2. Over three months, the average rate would be 2 * 3 = 6. Now, we can use the Poisson distribution to calculate the probability.
The probability of observing exactly k events in a Poisson distribution with an average rate of λ is given by the formula:
�
(
�
=
�
)
=
(
�
(
−
�
)
∗
�
�
)
/
�
!
P(X=k)=(e
(
−λ)∗λ
k
)/k!
Where:
�
(
�
=
�
)
P(X=k) is the probability of observing exactly k events.
�
e is the base of the natural logarithm, approximately 2.71828.
�
λ is the average rate.
�
k is the number of events.
In your case, you want to find the probability of at least 5 power outages, which means you need to find
�
(
�
≥
5
)
P(X≥5). To do this, you can calculate the probabilities for
�
=
0
,
1
,
2
,
3
,
4
k=0,1,2,3,4 and subtract the sum from 1.
�
(
�
≥
5
)
=
1
−
(
�
(
�
=
0
)
+
�
(
�
=
1
)
+
�
(
�
=
2
)
+
�
(
�
=
3
)
+
�
(
�
=
4
)
)
P(X≥5)=1−(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4))
Let's calculate this step by step:
Calculate the average rate for 3 months:
�
=
6
λ=6.
Calculate each individual term:
�
(
�
=
0
)
P(X=0):
�
(
�
=
0
)
=
(
�
(
−
6
)
∗
6
0
)
/
0
!
=
�
(
−
6
)
≈
0.00248
P(X=0)=(e
(
−6)∗6
0
)/0!=e
(
−6)≈0.00248
�
(
�
=
1
)
P(X=1):
�
(
�
=
1
)
=
(
�
(
−
6
)
∗
6
1
)
/
1
!
=
6
∗
�
(
−
6
)
≈
0.01488
P(X=1)=(e
(
−6)∗6
1
)/1!=6∗e
(
−6)≈0.01488
�
(
�
=
2
)
P(X=2):
�
(
�
=
2
)
=
(
�
(
−
6
)
∗
6
2
)
/
2
!
=
18
∗
�
(
−
6
)
≈
0.04464
P(X=2)=(e
(
−6)∗6
2
)/2!=18∗e
(
−6)≈0.04464
�
(
�
=
3
)
P(X=3):
�
(
�
=
3
)
=
(
�
(
−
6
)
∗
6
3
)
/
3
!
=
36
∗
�
(
−
6
)
≈
0.08928
P(X=3)=(e
(
−6)∗6
3
)/3!=36∗e
(
−6)≈0.08928
�
(
�
=
4
)
P(X=4):
�
(
�
=
4
)
=
(
�
(
−
6
)
∗
6
4
)
/
4
!
=
54
∗
�
(
−
6
)
≈
0.13392
P(X=4)=(e
(
−6)∗6
4
)/4!=54∗e
(
−6)≈0.13392
Now, calculate
�
(
�
≥
5
)
P(X≥5):
�
(
�
≥
5
)
=
1
−
(
0.00248
+
0.01488
+
0.04464
+
0.08928
+
0.13392
)
P(X≥5)=1−(0.00248+0.01488+0.04464+0.08928+0.13392)
�
(
�
≥
5
)
≈
1
−
0.28520
≈
0.7148
P(X≥5)≈1−0.28520≈0.7148
So, the probability of having at least 5 power outages in any three-month period is approximately 0.715 (rounded to three decimal places).