The correct answer is Option (1) →
|
$X$
|
0
|
1
|
2
|
|
$P(X)$
|
$\frac{188}{221}$
|
$\frac{32}{221}$
|
$\frac{1}{221}$
|
Let the random variable X be defined as the number of aces in a draw of 2 cards simultaneous, then X can take the values 0, 1, 2. The corresponding probabilities are:
$P(X = 0)$ = P(drawing no ace) = P(drawing both non-ace cards)
$=\frac{{^{48}C}_2}{{^{52}C}_2} =\frac{48 × 47}{1×2}×\frac{1×2}{52×51}=\frac{188}{221}$
$P(X = 1)$ = P(drawing one ace and one non-ace card)
$=\frac{{^4C}_1×{^{48}C}_1}{{^{52}C}_2}=\frac{4}{1}×\frac{48}{1}×\frac{1×2}{52×51}=\frac{32}{221}$
$P(X = 2)$ = P(drawing both aces) = $\frac{{^4C}_12}{{^{52}C}_2}=\frac{4×3}{1×2}×\frac{1×2}{52×51}=\frac{1}{221}$
|
$X$
|
0
|
1
|
2
|
|
$P(X)$
|
$\frac{188}{221}$
|
$\frac{32}{221}$
|
$\frac{1}{221}$
|
| 
