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

The Correct Option is D

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.