The outcomes of rolling fair six sided die is
. The prime numbers in the above outcomes are
. The probability of not rolling a prime number is
.
Each trial is independent.
In the given experiment of 100 trials, the binomial probability model can be used to predict the number of occurrence of non-prime numbers (X). The probability mass function is
. Since 

The tenth in 149.95 is 9. To round it to the nearest tenth, use the hundredth. The hundredth is 5. When rounding, always round up when it is a 5. 95 should be rounded to 100, but it is a decimal. 100 as a decimal is equal to 1. 149 + 1 = 150.
To find the volume of most objects the formula is length x width x height. So in order to find the volume you need to multiply the data.
1/2 x 1/6 x 1/3= 1/36
The volume is 1/36
Complete Question
The Brown's Ferry incident of 1975 focused national attention on the ever-present danger of fires breaking out in nuclear power plants. The Nuclear Regulatory Commission has estimated that with present technology there will be on average, one fire for every 10 years for a reactor. Suppose that a certain state has two reactors on line in 2020 and they behave independently of one another. Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that by 2030 at least two fires will have occurred at these reactors?
Answer:
The value is 
Step-by-step explanation:
From the question we are told that
The rate at which fire breaks out every 10 years is
Generally the probability distribution function for Poisson distribution is mathematically represented as

Here x represent the number of state which is 2 i.e 
Generally the probability that by 2030 at least two fires will have occurred at these reactors is mathematically represented as

=> ![P(x_1 + x_2 \ge 2 ) = 1 - [P(x_1 + x_2 = 0 ) + P( x_1 + x_2 = 1 )]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5BP%28x_1%20%2B%20x_2%20%3D%200%20%29%20%2B%20P%28%20x_1%20%2B%20x_2%20%3D%201%20%29%5D)
=> ![P(x_1 + x_2 \ge 2 ) = 1 - [ P(x_1 = 0 , x_2 = 0 ) + P( x_1 = 0 , x_2 = 1 ) + P(x_1 , x_2 = 0)]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5B%20P%28x_1%20%20%3D%200%20%2C%20%20x_2%20%3D%200%20%29%20%2B%20P%28%20x_1%20%3D%200%20%2C%20x_2%20%3D%201%20%29%20%2B%20P%28x_1%20%2C%20x_2%20%3D%200%29%5D)
=> 
=> ![P(x_1 + x_2 \ge 2 ) = 1 - \{ [ \frac{1^0}{ 0! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]] )+ ( [ \frac{1^1}{1! } * e^{-1}] * [[ \frac{1^1}{ 1! } * e^{-1}]] ) + ( [ \frac{1^1}{ 1! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]]) \}](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5C%7B%20%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%20%29%2B%20%28%20%5B%20%5Cfrac%7B1%5E1%7D%7B1%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E1%7D%7B%201%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%20%29%20%2B%20%28%20%5B%20%5Cfrac%7B1%5E1%7D%7B%201%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%29%20%5C%7D)
=> ![P(x_1 + x_2 \ge 2 )= 1- [[0.3678 * 0.3679] + [0.3678 * 0.3679] + [0.3678 * 0.3679] ]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%3D%201-%20%5B%5B0.3678%20%20%2A%200.3679%5D%20%2B%20%5B0.3678%20%20%2A%200.3679%5D%20%2B%20%5B0.3678%20%20%2A%200.3679%5D%20%20%5D)

1440
because yea
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