Probability question with a radio competition I'm quite new to statistics and I'm going through a few exam questions but I am a bit stuck on this one:
A radio station held a competition where contestants were invited to pick a
number from 1 to 50. If a contestant picked the ‘winning’ number they won a
trip to Vegas. The station picked a new ‘winning’ number at random each time
a new contestant played the game. The radio station allowed five contestants
to play every day over the course of one week.

    i. Compute the probability that the station will have to pay out for exactly 
one Vegas trip.
    ii. Compute the probability that the station will have to pay out for more
than 2 Vegas trips.
    iii. Suppose that the station decides to run the competition for an entire year
giving 5 × 365 = 1825 contestants. Find the approximate probability that
the number of Vegas trips they will have to pay for is greater than 50.

I'm not sure how to go about calculating this one and as I stated above I'm new to statistics so I'm not very good at it. Any help would be really appreciated
 A: 
Compute the probability that the station will have to pay out for exactly 
  one trip:

$$\sum\limits_{n=1}^{1}\binom{35}{n}\cdot\left(\frac{1}{50}\right)^{n}\cdot\left(1-\frac{1}{50}\right)^{35-n}$$

Compute the probability that the station will have to pay out for more
  than $2$ trips:

$$\sum\limits_{n=3}^{35}\binom{35}{n}\cdot\left(\frac{1}{50}\right)^{n}\cdot\left(1-\frac{1}{50}\right)^{35-n}$$
A: In the first two parts, the number $X$ of contestants winning trips is 
$X \sim Binom(35, 1/5).$ Here are partial PDF and CDF 4-place tables of that
distribution from R. Compare
your computed answers from the binomial formula with appropriate entries.
 n = 35;  p = 1/50;  i = 0:5
 pdf = round(dbinom(i, n, p), 4)
 cdf = round(pbinom(i, n, p), 4)
 cbind(i, pdf, cdf)
      i    pdf    cdf
      0 0.4931 0.4931
      1 0.3522 0.8453       
      2 0.1222 0.9675
      3 0.0274 0.9949      
      4 0.0045 0.9994
      5 0.0006 0.9999

For the last part you have $Y \sim Binom(1825, 1/50),$ and you
seek $P(Y > 50).$ Results below are for the exact binomial value
(perhaps too tedious to compute by hand), and Poisson and normal
approximations.
 n = 1825; p = 1/50;  mu = n*p;  sg = sqrt(n*p*(1-p))
 1 - pbinom(50, n, p)
 ## 0.01257754
 1 - ppois(50, mu)
 ## 0.01338316
 1 - pnorm(50.5, mu, sg)
 ## 0.009620645

The figure below shows the PDF of $Binom(1825, 1/50)$ (black bars),
and the Poisson approximation (blue open circles) and the
best-fitting normal curve (red curve).

