A young baseball fan wants to collect a complete set of 262 baseball cards. The baseball cards are available in a completely random fashion, one per package of chewing gum.
How many boxes of chewing gum does the fan need to buy in order to have a full set with probability ≥ $0.99$?
I was told I need more info to solve this problem. However, this is all the info given. Is this missing data?
As lulu noted in a commented, this is the coupon collector's problem. The probability of having a complete set of $m$ coupons after drawing $n$ coupons is
$$ \def\stir#1#2{\left\{#1\atop#2\right\}} \frac{m!}{m^n}\stir nm\;, $$
where $\stir nm$ is a Stirling number of the second kind. For a derivation of this probability, see Probability distribution in the coupon collector's problem.
You have $m=262$ and want the probability to be at least $0.99$. Since we can't easily solve for $m$, we should try to get a good estimate so we don't have to compute too many Stirling numbers. The number of coupons you need to draw to have all coupons with probability $0.99$ should be close to the number you need to draw to make the expected number of undrawn coupons $0.01$. So
$$ m\left(1-\frac1m\right)^n\approx0.01\;, $$
and solving for $n$ yields
$$ n\approx\frac{\log0.01-\log m}{\log\left(1-\frac1m\right)}\approx2660.37\;, $$
and indeed calculating the exact probability for the values adjacent to $n=2660$ shows that, as Byron found, the probability $0.99$ is first reached at this value.