A die is thrown three times. Let `X` be the number of twos seen. Find the expectation of `X` .

Explanation

We have, X=number of twos seen
So, on throwing a die three times, we will have X=0,1,2,3.
`therefore P(X=0)=P_(("not" 2))cdotP_(("not" 2))cdotP_(("not" 2))=5/6cdot5/6cdot5/6=125/216`
`P(X=1)=P_(("not" 2))cdotP_(("not" 2))cdotP_((2))+P_(("not" 2))cdotP_((2))cdotP_(("not" 2))+P_((2))cdotP_(("not"2))cdotP_(("not" 2))`
`=5/6cdot5/61/6+5/6cdot1/6cdot5/6+1/6cdot5/6cdot5/6=25/36cdot3/6=25/72`
`P(X=2)=P_(("not" 2))cdotP_((2))cdotP_((2))+P_((2))cdotP_((2))cdotP_(("not" 2))+P_((2))`
`=5/6cdot1/6cdot1/6+1/6cdot1/6cdot5/6+1/6cdot5/6cdot1/6`
`=1/36cdot[15/6]=15/216`
`P(X=3)=P_((2))cdotP_((2))cdotP_((2))=1/6cdot1/6cdot1/6=1/216` ltbr gtWe know that, `E(X)=sumXP(X)=0cdot125/216+1cdot25/72+2cdot15/216+3cdot1/216`
`=(75+30+3)/216=108/216=1/2`