please give a better explanation of what you want to be answered.
Answer:
$$\begin{align*}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Explanation:
\eqalign{
P(Y-X=m\mid Y\gt X)
&=\sum_kP(Y-X=m,X=k\mid Y\gt X)\cr
&=\sum_kP(Y-X=m\mid X=k,Y\gt X)\,P(X=k\mid Y>X)\cr
&=\sum_kP(Y-k=m\mid Y\gt k)\,P(X=k\mid Y\gt X)\cr
}
P(Y-X=m | Y > X) &= \sum_{k} P(Y-X=m, X=k | Y > X) \\ &= \sum_{k} P(Y-X=m | X=k, Y > X) P(X=k | Y > X) \\ &= \sum_{k} P(Y-k=m | Y > k) P(X=k | Y > X).\end{split}$$
Answer:
View Image
Explanation:
You didn't provide me a picture of the opamp.
I'm gonna assume that this is an ideal opamp, therefore the input impedance can be assumed to be ∞ . This basically implies that...
- no current will go in the inverting(-) and noninverting(+) side of the opamp
- V₊ = V₋ , so whatever voltage is at the noninverting side will also be the voltage at the inverting side
Since no current is going into the + and - side of the opamp, then
i₁ = i₂
Since V₊ is connected to ground (0V) then V₋ must also be 0V.
V₊ = V₋ = 0
Use whatever method you want to solve for v_out and v_in then divide them. There's so many different ways of solving this circuit.
You didn't give me what the input voltage was so I can't give you the entire answer. I'll just give you the equations needed to plug in your values to get your answers.