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

Answer

2/13

Explanation

Let `X` represents the number of aces.
Therefore, the possible values of `X` are 0,1 are 2.
Since there are 4 aces in a pack of 52 cards, so there are 48 other cards.
`:. P(X=0)=P` (not getting any ace)
`=(.^(48)C_(2))/(.^(52)C_(2))=(48xx47)/(52xx51)=188/221`
`P(X=1)=P` (getting one ace and one other card)
`=(.^(4)C_(1)xx.^(48)C_(1))/(.^(52)C_(2))=(4xx48)/((52xx51)/(2xx1))=32/221`
and `P(X=2)=P` (getting 2 aces)
`=(.^(4)C_(2))/(.^(52)C_(2))=(4xx3)/(52xx51)=1/221`
Therefore the probability distribution of `X` is

`:.E(X)=sumXP(X)`
`=0xx188/221+1xx32/221+2xx1/221`
`=34/221=2/13`