Solving Binomial probability question- what's wrong with my solution? 
A company establishes a fund of $120$ from which it wants to pay an amount, $C$, to any of its $20$ employees who achieve a high-performance level during the coming year. each employee has a $2\%$ chance of achieving a high-performance level during the coming year, independent of any other employees. Determine the maximum value of $C$ for which the probability is less than $1\%$ that the fund will be inadequate to cover all payments for high performance.
Multiple-Choice Options: $24, 30, 40, 60, 120$.

My attempt:
Let $X$ be the RV denoting the number of employees that need to be paid the amount $C$. $X \sim Bin(20, 0.02)$. We need to find $P(X>120/C)<0.01$.
Note that $P(X= 0) = 0.667$, $P(X= 1) = 0.272$, $P(X= 2) = 0.052$, $P(X= 3) = 0.006$. Since $P(X= 3)<0.01$, we obtain $$3>\frac{120}{C} \implies C>40$$
I chose the option for $40$, however, the correct answer is $60$. Now, since my answer said $C>40$, I should have chosen $60$ or $120$, but I suspect that the solution has major flaws. Can someone please review my solution and let me know where I went wrong? Thanks.
 A: The fund will be insufficient if $X$ is "too large."  We want to limit the probability of this.  This suggests finding some integer $c$ such that $$\Pr[X > c] < 0.01. \tag{1}$$ Since $X \sim \operatorname{Binomial}(n = 20, p = 0.02)$, we calculate a table:
$$\begin{array}{c|c|c|c}
x & \Pr[X = x] & \Pr[X \le x] & \Pr[X > x] \\
\hline
0 & 0.667608 & 0.667608 & 0.332392 \\
1 & 0.272493 & 0.940101 & 0.059899 \\
2 & 0.0528303 & 0.992931 & 0.00706869 \\
3 & 0.00646901 & 0.9994 & 0.000599679
\end{array}$$
The third column is computed by successively adding up entries in the second column, and the fourth column is $1$ minus the third column.  We observe that the first value of $x$ for which the desired inequality $(1)$ is satisfied is $x = 2$; therefore, this is our value of $c$.  Hence the probability of having more than $2$ high-performing employees is less than $0.01$, and if the award is $C$, the company can make $C = 60$ without having the probability of insufficient funds exceed $0.01$, since the probability of three or more high-performing employees is less than $0.01$.

Your error is in concluding $c = 3$ rather than $c = 2$.  But there is a more fundamental error, which is that you did not compute cumulative probabilities.  Instead, you computed individual probabilities and stopped when $\Pr[X = c] < 0.01$.  This is not correct, as the inequality $(1)$ above explains.
A: From your  $P(X= 0) = 0.668$, $P(X= 1) = 0.272$, $P(X= 2) = 0.053$, $P(X= 3) = 0.006$ (with slightly better rounding)
you can find $P(X> 0) = 0.332$, $P(X> 1) = 0.060$, $P(X> 2) = 0.007$, $P(X> 3) = 0.001$.
So there is a probability of less than $1\%$ that there are more than $2$ high-performers, but more than $1\%$ that there is more than $1$ high-performer,
and thus you need the fund of $120$ to be able to pay out to up to two high performers at a potential cost of $2C$.
$2C \le 120$ implies $C \le 60$.
