<s>[INST] Four cards are drawn without replacement. What is the probability of drawing at least two kings? Four cards are drawn without replacement. What is the probability of drawing at least two kings?
is the below my answer correct or not?!
since cards are drawn without replacement, (4/52)(3/51)(48/50)*(47/49)=1/240=0.0042
1-0.0042=0.9958
Probability of drawing AT LEAST 2 kings is 0.9958.
 [/INST] There are $\binom{4 \cdot 13}{4} = \binom{52}{4} = 270725$ cases of "drawing four cards", the ways to get $2$ out of $4$ kings is $\binom{4}{2}$, to get $3$ is $\binom{4}{3}$, for $4$ it is $\binom{4}{4}$. In all:
$$\binom{4}{2} + \binom{4}{3} + \binom{4}{4}
    = 11$$
The probability is thus $\frac{11}{270725} = 4.06316 \cdot 10^{-5}$.
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