Question

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. What is the value of E(X)?

• A. $\frac{37}{221}$
• B. $\frac{5}{13}$
• C. $\frac{1}{13}$
• D. $\frac{2}{13}$

Answer 1

Answer: (D)

$\frac{2}{13}$

Answer 2

Let X denote the number of aces obtained. Therefore, X can take any of the values of 0, 1, or 2. In a deck of 52 cards, 4 cards are aces. Therefore, there are 48 non-ace cards.

∴ P (X = 0) = P (0 ace and 2 non-ace cards) = $\frac{^4C_0*^48C_2}{^52C_2}=\frac{1128}{1326}$

P (X = 1) = P (1 ace and 1 non-ace cards) =$\frac{^4C_0*^48C_2}{^52C_2}=\frac{192}{1326}$

P (X = 2) = P (2 ace and 0 non- ace cards) = $\frac{^4C_0*^48C_2}{^52C_2}=\frac{6}{1326}$

Thus, the probability distribution is as follows.

X	0	1	2
P(x)	$\frac{1128}{1326}$	$\frac{192}{1326}$	$\frac{6}{1326}$

Then, E(X) = $\sum p_ix_i$

= 0*$\frac{1128}{1326}$+1*$\frac{192}{1326}$+2*$\frac{6}{1326}$

=$\frac{204}{1326}$

=$\frac{2}{13}$

Therefore, the correct answer is D.