Question

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is (A) (B) (C) (D)

Answer 1

It is given that X is the number of aces obtained. The possible values of X are 0, 1 or 2.

Total number of aces in a deck of cards is 4 and number of non-ace cards is 48.

The probability of getting two non-ace cards is,

P( X=0 )=P( 0 ace and 2 non-ace cards ) = 4 C 0 × C 48 2 52 C 2 = 4! 0!( 4−0 )! × 48! 2!( 48−2 )! 52! 2!( 52−2 )! = 4! 4! × 48! 2!( 46 )! 52! 2!( 50 )!

Solving further,

P( X=0 )= 1× 48×47×46! 2!( 46 )! 52×51×50! 2!( 50 )! = 24×47 26×51 = 1128 1326

The probability of getting one ace and one non-ace card is,

P( X=1 )=P( 1 ace and 1 non-ace cards ) = C 4 1 × C 48 1 52 C 2 = 4! 1!( 4−1 )! × 48! 1!( 48−1 )! 52! 2!( 52−2 )! = 4! 3! × 48! 1!( 47 )! 52! 2!( 50 )!

Solving further,

P( X=1 )= 4× 48×47! ( 47 )! 52×51×50! 2!( 50 )! = 4×48 26×51 = 192 1326

The probability of getting two aces and zero non-ace cards is,

P( X=2 )=P( 2 ace and 0 non-ace cards ) = 4 C 2 × C 48 0 52 C 2 = 4! 2!( 4−2 )! × 48! 0!( 48−0 )! 52! 2!( 52−2 )! = 4! 2!×2! × 48! 0!( 48 )! 52! 2!( 50 )!

Solving further,

P( X=1 )= 4×3 2 × 48! ( 48 )! 52×51×50! 2!( 50 )! = 6×1 26×51 = 6 1326

The distribution of probability for different cases of X is as follows,

X	P( X )
0	1128 1326
1	192 1326
2	6 1326

The value of E( X ) is calculated as,

E( X )= ∑ i=1 n p i x i =0× 1128 1326 +1× 192 1326 +2× 6 1326 = 204 1326 = 2 13

Thus, the correct option is (D).