Let 
denote the value on the 
-th drawn ball. We want to find the expectation of 
, which by linearity of expectation is
![E[S]=E\left[\displaystyle\sum_{i=1}^5B_i\right]=\sum_{i=1}^5E[B_i]](https://tex.z-dn.net/?f=E%5BS%5D%3DE%5Cleft%5B%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E5B_i%5Cright%5D%3D%5Csum_%7Bi%3D1%7D%5E5E%5BB_i%5D)
(which is true regardless of whether the 
are independent!)
At any point, the value on any drawn ball is uniformly distributed between the integers from 1 to 10, so that each value has a 1/10 probability of getting drawn, i.e.
and so
![E[X_i]=\displaystyle\sum_{i=1}^{10}x\,P(X_i=x)=\frac1{10}\frac{10(10+1)}2=5.5](https://tex.z-dn.net/?f=E%5BX_i%5D%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7B10%7Dx%5C%2CP%28X_i%3Dx%29%3D%5Cfrac1%7B10%7D%5Cfrac%7B10%2810%2B1%29%7D2%3D5.5)
Then the expected value of the total is
![E[S]=5(5.5)=\boxed{27.5}](https://tex.z-dn.net/?f=E%5BS%5D%3D5%285.5%29%3D%5Cboxed%7B27.5%7D)
Answer:
Option 3
Step-by-step explanation:
The coordinates match it
There are 81 girls to every 72 boys in total, which can be represented by 81/72, which put into decimal form, is 1.125. If you then take the number of boys in the group, 16, and multiply it by this number (because the problem states that the ratio is constant) you can find the number of girls in the group.
16*1.125=18
So there are 18 girls in a group with 16 boys.
Answer:
Step-by-step explanation:
42-42(2/7)
42-12
30
There are 30 options with no chocolate.
No it is not the worst subject
