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.
Best Answer
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 $$1 - \sum_{x=0}^{2}{52\choose x}(0.02^{x})(0.98^{52-x})$$