The problem is such that how to find expected value if variable y has the following distributn( kind of discrete)
$$p(y=0)=0.2$$ $$p(y=1)=0.3$$ $$p(y=2)=0.3$$ $$p(y>2)=0.2 $$
So i was confused by last term , i think we wouldnt consider cases when $y=3,4,5, \ldots$ so i am aware that it is possible to rewrite $p(y>2)$ as $1-p(y=1)-p(y=0)$ but then probabilities do not coincide. So how to find EY?
Let $X\sim Poi(\lambda)$. Then $P(X>2)=1-P(X=2)-P(X=1)-P(X=0)$.
$$ =1-e^{-\lambda}(\frac{\lambda^2}2+\lambda+1)=0.2$$
This equation cannot be solved algebraically. So use a calculator. Then
$$E(Y)=0\cdot 0.2+1\cdot 0.3+2\cdot 0.3+\sum_{x=3}^{\infty} x\cdot e^{-1.53504}\cdot \frac{1.53504^x}{x!}=1.597$$
The information that $Y \sim \mathsf{Pois}(\lambda)$ is crucial. Then $P(Y = 0) = 0.2e^{-\lambda}$ can be solved to get $\lambda \approx 1.61$ (as @WaveX suggests; take logs). So $E(Y) \approx 1.61.$
But the probabilities in your problem seem to have been severely rounded. Using the Poisson PDF with $\lambda = 1.61,$ I get the following probabilities, where the last line is for $P(Y \ge 3):$
y P(Y=y) Rounded
0 0.1998876 0.2
1 0.3218191 0.3
2 0.2590643 0.3
3+ 0.2192290 0.2
Without the assumption that $Y$ has a Poisson distribution, you can say that
$$E(Y) \ge 0(.2) + 1(.3) + 2(.3) + 3(.2) = 1.5,$$
but you don't have enough information to get an exact value. Using the value $\lambda = 1.5$ with the Poisson PDF, the table above becomes:
cbind(x,pdf)
y P(Y-y) Rounded
0 0.2231302 0.2
1 0.3346952 0.3
2 0.2510214 0.3
3+ 0.1911532 0.2
So the much-rounded version of the distribution is the same as for $\lambda = 1.61.$
I honestly don't know what answer you should give to this flawed exercise. You can give the first argument above with answer $E(Y) \approx 1.61;$ you can give the second argument with answer $E(Y) \approx 1.5;$ or you can give both along with a polite suggestion that the problem is seriously flawed and you did what you could.
Below is a plot that illustrates the difference between the distributions $\mathsf{Pois}(1.61)$ and $\mathsf{Pois}(1.5).$
Adding up to Bruce's response, you could take the given probabilities as observed data distribution and not real probabilities, and use the four of them, together with the true Poisson distribution you already know, to find a maximum likelihood estimator for $\lambda$.
Here is how: assume the given probabilities correspond to a data sample of size $N$, ($y_1,\dots,y_N$), such that there are $n_0$ observations of 0 arrivals ($y_k=0$), $n_1$ of 1 arrival, $n_2$ of 2 arrivals, and $n_3$ of more than 2 arrivals. So $n_0+n_1+n_2+n_3=N$ and $\frac{n_0}{N}=0.2$, $\frac{n_1}{N}=0.3$, $\frac{n_2}{N}=0.3$, $\frac{n_3}{N}=0.2$.
The likelihood function of the parameter $\lambda$ given the data is: $$L(\lambda|y_1,\dots,y_k)=\prod_{k=1}^N P(Y=y_k|\lambda)=P(Y=0)^{n_0}P(Y=1)^{n_1}P(Y=2)^{n_2}P(Y>2)^{n_3}.$$
(I dropped the condition on $\lambda$ to simplify the notation.)
We take logarithm that preserves concavity and so does not change which $\lambda$ will give the maximum likelihood:
$$logL(\lambda|y_1,\dots,y_N)=n_0\log\left(P(Y=0)\right)+n_1\log\left(P(Y=1)\right)+n_2\log\left(P(Y=2)\right)+n_3\log\left(P(Y>2)\right).$$
Now we substitute each $P(Y=y)$ by its corresponding expression on $\lambda$ using the Poisson distribution formula:
$$logL(\lambda)=n_0\log\left(e^{-\lambda}\right)+n_1\log\left(e^{-\lambda}\lambda\right)+n_2\log\left(e^{-\lambda}\frac{\lambda^2}{2}\right)+n_3\log\left(1-e^{-\lambda}\left(1+\lambda+\frac{\lambda^2}{2}\right)\right).$$
Dividing by a positive constant does not change where the maximum of the function is located, so we divide both sides by $N$ (and substitute the $\frac{n_k}{N}$ by their corresponding probabilities given in the problem):
$$\frac{logL(\lambda)}{N}=0.2\log\left(e^{-\lambda}\right)+0.3\log\left(e^{-\lambda}\lambda\right)+0.3\log\left(e^{-\lambda}\frac{\lambda^2}{2}\right)+0.2\log\left(1-e^{-\lambda}\left(1+\lambda+\frac{\lambda^2}{2}\right)\right).$$
We take derivative and equate it to zero to solve for $\lambda$ and we find the MLE estimator given the data is:
$$\hat{\lambda}=1.60187.$$
Finally, for a Poisson distribution with arrival rate $\lambda$, $E(Y)=\lambda$, so you obtain the maximum likelihood estimator for the mean is
$$\hat{\mu}=1.60187.$$