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**Soroban** Hello, khor!

The second one has an incredibly long and tedious solution.

Since it depends on how many diamonds were removed,

. . we must consider all possible cases.

Suppose no $\displaystyle \diamondsuit$ is removed. . This probability is: .$\displaystyle \frac{{39\choose21}}{{52\choose21}} $

The deck contains: 13 $\displaystyle \diamondsuit$s and 18 Others.

The probability that no $\displaystyle \diamondsuit$ is drawn is: .$\displaystyle \frac{{18\choose2}}{{31\choose2}} $

The probability that no $\displaystyle \diamondsuit$ is removed and at least one $\displaystyle \diamondsuit$ is drawn is: .$\displaystyle \frac{{39\choose21}}{{52\choose21}}\left(1 - \frac{{18\choose2}}{{31\choose2}} \right) $

Suppose 1 $\displaystyle \diamondsuit$ is removed. . This probability is: .$\displaystyle \frac{{13\choose1}{39\choose20}}{{52\choose21}} $

The deck contains 12 $\displaystyle \diamondsuit$s and 19 Others.

The probability that no $\displaystyle \diamondsuit$ is drawn is: .$\displaystyle \frac{{19\choose2}}{{31\choose2}} $

The probabiity that 1 $\displaystyle \diamondsuit$ is removed and at least one $\displaystyle \diamondsuit$ is drawn is: .$\displaystyle \frac{{13\choose1}{39\choose20}}{{52\choose21}}\le ft(1 - \frac{{19\choose2}}{{31\choose2}}\right) $

. . . $\displaystyle \vdots$

We must extend this list up to:

Suppose 11 $\displaystyle \diamondsuit$s are removed. . This probability is: .$\displaystyle \frac{{13\choose11}{39\choose10}}{{52\choose21}}

$

The deck contains 2 $\displaystyle \diamondsuit$s and 29 Others.

The probability that no $\displaystyle \diamondsuit$ is drawn is: .$\displaystyle \frac{{29\choose2}}{{31\choose2}} $

The probability that 11 $\displaystyle \diamondsuit$s are removed and at least one $\displaystyle \diamondsuit$ is drawn is: .$\displaystyle \frac{{13\choose11}{39\choose10}}{{52\choose21}} \left(1 - \frac{{29\choose2}}{{31\choose2}}\right) $

The final probability is the sum of these twelve probabilities.

You go on ahead . . . I'll wait in the car.

.