The bag contains,
Red (R) marbles is 9, Green (G) marbles is 7 and Blue (B) marbles is 4,
Total marbles (possible outcome) is,

Let P(R) represent the probablity of picking a red marble,
P(G) represent the probability of picking a green marble and,
P(B) represent the probability of picking a blue marble.
Probability , P, is,


Probablity of drawing a Red marble (R) and then a blue marble (B) without being replaced,
That means once a marble is drawn, the total marbles (possible outcome) reduces as well,

Hence, the best option is G.
Answer:
![E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}](https://tex.z-dn.net/?f=E%28X%29%3D%20n%20%5Cint_%7B0%7D%5E1%20x%5En%20dx%20%3D%20n%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7D-%20%5Cfrac%7B0%7D%7Bn%2B1%7D%5D%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)
Step-by-step explanation:
A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".
We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).
If we select a value
we want this:

And we can express this like that:
for each possible i
We assume that the random variable
are independent and
from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

And then cumulative distribution would be expressed like this:



For each value
we can find the dendity function like this:

So then we have the pdf defined, and given by:
and 0 for other case
And now we can find the expected value for the random variable X like this:

![E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}](https://tex.z-dn.net/?f=E%28X%29%3D%20n%20%5Cint_%7B0%7D%5E1%20x%5En%20dx%20%3D%20n%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7D-%20%5Cfrac%7B0%7D%7Bn%2B1%7D%5D%3D%5Cfrac%7Bn%7D%7Bn%2B1%7D)
Answer:
AP = 22
Step-by-step explanation:
In a triangle, the centroid divides the median in the ratio 2:1.
It is given that AD is the median and AD = 33
It is also given that P is the centroid on the median AD.
Therefore, P divides AD in the ratio 2:1.

AP = 2k and PD = k
AD = 33
AP + PD = 33
2k + k = 33
3k = 33
k = 11
So, AP = 2k = 2(11) = 22