The variable is how many tanks there are because he wants to know how many fish there are in the whole aquarium and we already know each tank contains five fishes but we just need to know how many tanks there are. :)
Divide 45.5 and 13
to get you 3.5
i hope i helped
To solve the question we use the following analogy:
Given that the odds of winning is a/b, then the probability is b/(a+b)
hence to solve the question we proceed as follows:
a]
P(E)=11/36
odds against E=(36-11)/11
simplifying this gives us:
25/11
b] P(E)=31/36
odds against E=(36-31)/31
simplifying this we get:
5/31
hence odds in favor of E=31/5
c]P(E)=32/36
odds in favor of E will be:
36/(36-32)
simplifying the above we get:
36/4
=9/1
d]
P(E)=30/36
odds against E will be:
(36-30)/36
simplifying this we obtain:
6/36
=1/5
e] P(E)=13/36
odds against E will be:
(36-13)/13
simplifying this we obtain:
23/13
Answer:
a) 
b) 
c) 
d) 
e) 
f) 
g) 
h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]
Step-by-step explanation:
For this case we know this:
with both Y and u random variables, we also know that:
![[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2](https://tex.z-dn.net/?f=%20%5Btex%5D%20E%28v%29%20%3D%200%2C%20Var%28v%29%20%3D1%2C%20E%28X%29%20%3D%201%2C%20Var%28X%29%3D2)
And we want to calculate this:
Part a

Using properties for the conditional expected value we have this:

Because we assume that v and X are independent
Part b

If we distribute the expected value we got:

Part c

Using properties for the conditional expected value we have this:

Because we assume that v and X are independent
Part d

If we distribute the expected value we got:

Part e

Part f

Part g

Part h
E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]