Answer:
![P(k) = \frac{\binom{N}{k} \binom{(M+N) - N}{r-k}}{\binom{M+N}{r}}](https://tex.z-dn.net/?f=P%28k%29%20%3D%20%5Cfrac%7B%5Cbinom%7BN%7D%7Bk%7D%20%5Cbinom%7B%28M%2BN%29%20-%20N%7D%7Br-k%7D%7D%7B%5Cbinom%7BM%2BN%7D%7Br%7D%7D)
Step-by-step explanation:
We can model the string as a hypergeometric distribution, as each bit has two possible values, 1 or 0, and the chance of a 1 or 0 changes with every bit, as there are a finite number M of 0's and N of 1's and every bit takes one of those values.
If M+N (total size of the string) >> r (number of trials), we could model it as a binomial distribution as the probability of a 1 or 0 wouldn't change in a significant amount with every bit, but as we don't know the magnitude of M+N and r, we follow up with hypergeometric distribution.
The distribution has the following formula for probability:
![P(k) = \frac{\binom{K}{k} \binom{N - K}{n-k}}{\binom{N}{n}}](https://tex.z-dn.net/?f=P%28k%29%20%3D%20%5Cfrac%7B%5Cbinom%7BK%7D%7Bk%7D%20%5Cbinom%7BN%20-%20K%7D%7Bn-k%7D%7D%7B%5Cbinom%7BN%7D%7Bn%7D%7D)
Where k is the number of sucesses, K is how many total sucess states are in the population, N is the population size and n is the number of draws.
For our case, a 1 would be a sucess, i.e. k the number of 1's we want to know the probability, N our total number of 1's, M+N the length of the string (population size) and we want to analyse what happens in the first r bits (number of draws):
![P(k) = \frac{\binom{N}{k} \binom{(M+N) - N}{r-k}}{\binom{M+N}{r}}](https://tex.z-dn.net/?f=P%28k%29%20%3D%20%5Cfrac%7B%5Cbinom%7BN%7D%7Bk%7D%20%5Cbinom%7B%28M%2BN%29%20-%20N%7D%7Br-k%7D%7D%7B%5Cbinom%7BM%2BN%7D%7Br%7D%7D)
Some examples of perpendicular lines are the Interstate how it runs side by side with a median in the middle the roads are perpendicular because a lot of them will cross one another in a sharp turn trust me I know, Some more are the studs in the walls of your house, the wheels on your truck, the frame rails on your go cart just look around you if the lines intersect at a right angle then there perpendicular.
For this case we have that by definition, the volume of a rectangular prism is given by:
![V = A_ {b} * h](https://tex.z-dn.net/?f=V%20%3D%20A_%20%7Bb%7D%20%2A%20h)
Where:
It is the area of the base
h: It is the height
So:
![A_ {b} = 8 * 4 = 32 \ in ^ 2\\h = 214 \ in](https://tex.z-dn.net/?f=A_%20%7Bb%7D%20%3D%208%20%2A%204%20%3D%2032%20%5C%20in%20%5E%202%5C%5Ch%20%3D%20214%20%5C%20in)
The volume of the prism is:
![V = 32 * 214 = 6848 \ in ^ 3](https://tex.z-dn.net/?f=V%20%3D%2032%20%2A%20214%20%3D%206848%20%5C%20in%20%5E%203)
Now, the volume of the cubes is:
![V = (14) ^ 3 = 2744 \ in ^ 3](https://tex.z-dn.net/?f=V%20%3D%20%2814%29%20%5E%203%20%3D%202744%20%5C%20in%20%5E%203)
Thus, we divide and find the number of cubes to fill the prism:
![n = \frac {6848 \ in ^ 3} {2744 \ in ^ 3}\\n = 2.5](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%20%7B6848%20%5C%20in%20%5E%203%7D%20%7B2744%20%5C%20in%20%5E%203%7D%5C%5Cn%20%3D%202.5)
ANswer:
n = 2.5
Isn't it 1/8?
I hope this helped i think i just simplified xD