The correct answer is Option (1) →
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$X$
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0
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1
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2
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$P(X)$
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$\frac{{^{48}C}_2}{{^{52}C}_2}$
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$\frac{{^4C}_1×{^{48}C}_1}{{^{52}C}_2}$
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$\frac{{^4C}_2}{{^{52}C}_2}$
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Total ways to choose 2 cards from 52: ${}^{52}\!C_{2}$
Case 1: X = 0
Ways to choose 2 non-kings from 48: ${}^{48}\!C_{2}$
$P(X = 0) = \frac{{}^{48}\!C_{2}}{{}^{52}\!C_{2}}$
Case 2: X = 1
Ways to choose 1 king from 4 and 1 non-king from 48: ${}^{4}\!C_{1} \cdot {}^{48}\!C_{1}$
$P(X = 1) = \frac{{}^{4}\!C_{1} \cdot {}^{48}\!C_{1}}{{}^{52}\!C_{2}}$
Case 3: X = 2
Ways to choose 2 kings from 4: ${}^{4}\!C_{2}$
$P(X = 2) = \frac{{}^{4}\!C_{2}}{{}^{52}\!C_{2}}$
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