Answer:
Step-by-step explanation:
a.
4 people can be selected in ways=12 P4=12×11×10×9=11880
b.
4 women can be selected in ways=7P4=7×6×5×4=840
c.
![P=\frac{840}{11880} =\frac{7}{99}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B840%7D%7B11880%7D%20%3D%5Cfrac%7B7%7D%7B99%7D)
Well, 10x4=40 because 10+10+10+10=40 and that means your answer must be 40 ones or you asked something else.
Answer:
![P(X\leq 1) = 0.331](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%20%3D%200.331)
Step-by-step explanation:
Given
Poisson Distribution;
Average rent in a week = 2.3
Required
Determine the probability of renting no more than 1 apartment
A Poisson distribution is given as;
![P(X = x) = \frac{y^xe^{-y}}{x!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7By%5Exe%5E%7B-y%7D%7D%7Bx%21%7D)
Where y represents λ (average)
y = 2.3
<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>
<em />
Using probability notations;
![P(X\leq 1) = P(X=0) + P(X =1)](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%20%3D%20P%28X%3D0%29%20%2B%20P%28X%20%3D1%29)
Solving for P(X = 0) [substitute 0 for x and 2.3 for y]
![P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7B2.3%5E0%20%2A%20e%5E%7B-2.3%7D%7D%7B0%21%7D)
![P(X = 0) = \frac{1 * e^{-2.3}}{1}](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7B1%20%2A%20e%5E%7B-2.3%7D%7D%7B1%7D)
![P(X = 0) = e^{-2.3}](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20e%5E%7B-2.3%7D)
![P(X = 0) = 0.10025884372](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%200.10025884372)
Solving for P(X = 1) [substitute 1 for x and 2.3 for y]
![P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7B2.3%5E1%20%2A%20e%5E%7B-2.3%7D%7D%7B1%21%7D)
![P(X = 1) = \frac{2.3 * e^{-2.3}}{1}](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7B2.3%20%2A%20e%5E%7B-2.3%7D%7D%7B1%7D)
![P(X = 1) =2.3 * e^{-2.3}](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D2.3%20%2A%20e%5E%7B-2.3%7D)
![P(X = 1) = 2.3 * 0.10025884372](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%202.3%20%2A%200.10025884372)
![P(X = 1) = 0.23059534055](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%200.23059534055)
![P(X\leq 1) = P(X=0) + P(X =1)](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%20%3D%20P%28X%3D0%29%20%2B%20P%28X%20%3D1%29)
![P(X\leq 1) = 0.10025884372 + 0.23059534055](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%20%3D%200.10025884372%20%2B%200.23059534055)
![P(X\leq 1) = 0.33085418427](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%20%3D%200.33085418427)
![P(X\leq 1) = 0.331](https://tex.z-dn.net/?f=P%28X%5Cleq%201%29%20%3D%200.331)
Hence, the required probability is 0.331
Each digit can be 0-9 ( 10 numbers)
there are 9 numbers in a social security number
multiply 10 nine times:
10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,000,000,000 ( 1 billion) possibilities.
the restriction is it can't be all zeros so subtract 1 from the total
1,000,000,000 - 1 = 999,999,999 combinations