Binomial Distribution - probability of winning 3 or more lottery prizes if you buy 1 ticket per week The question asks:

Suppose that in a weekly lottery you have probability .02 of winning a prize with a single
  ticket. If you buy 1 ticket per week for 52 weeks, what is the probability that you win 3 or more prizes?

I realized that the probability function of this problem would can be represented by a binomial distribution such that if we let $x =$ number of prizes won then the probability that I win $3$ or more prizes can be represented by the cumulative distribution function $$P(X\ge3) = \binom{52}{x}(0.02^x)(0.98^{52-x})$$
To find $x$ I realized that to satisfy the condition that $X\ge x$ I needed $x \in \{3,4,5...,52\}$ since there are $50$ possible numbers that $x$ can be I let $x = 50$. The answer should be $0.0859$ but I cannot seem to get that answer.
 A: First, notice that you mis-wrote the Binomial distribution. For some $x$, you should have ${52\choose x}(0.02^{x})(0.98^{52-x})$, and not the power of $x$ for both your successes and failures (evidently).
Then much easier to calculate is realizing that this is equivalent to 
$$ 1 - P(\text{win less than 3 prizes})$$
which is given by $$1 - \sum_{x=0}^{2}{52\choose x}(0.02^{x})(0.98^{52-x})$$
A: The correct formula for $P(X)$ is
$$
P(X = x) = \binom{52}{x}(0.02)^x(0.98)^{52-x}.
$$
Then the formula for $P(X \ge 3)$ is,
$$
P(X \ge 3) = \sum_{x = 3}^{52} \binom{52}{x}(0.02)^x(0.98)^{52-x}.
$$
Of course, if you don't have a computer available, it's going to be much more feasible to compute $1 - P(X \ge 3) = P(X \le 2)$, as Donkey Kong suggests.
A: You have to use the cummulative distribution.
$P(X \geq 3)=\sum_{k=3}^{52} {52 \choose k} \cdot 0.02^k\cdot0.98^{52-k}$
If you calculate like this then you have to sum up 50 terms. A better way is to use the converse probability.
$1-P(X \leq 2)=1-\sum_{k=0}^{2} {52 \choose k} \cdot 0.02^k\cdot 0.98^{52-k}\approx 0.0859$
A: As you said, this can be modeled by binomial distribution.
The answer should be 
$$
 P(X \ge 3) = 1 - (P( X = 0) + P(X = 1) + P(X = 2))
$$
$$
P(X = 0) = { 52 \choose 0 } 0.02^0 *  0.98^{52} = 0.34975
$$
$$
P(X = 1) = { 52 \choose 1 } 0.02^1 *  0.98^{51} = 0.371162
$$
$$
P(X = 2) = { 52 \choose 2 } 0.02^2 *  0.98^{50} = 0.19315
$$
Therefore,
$$
 P(X \ge 3) = 1 - (0.34975 + 0.371162 + 0.19315) = 0.085938
$$
