On a normal standard die one of the 21 dots from any one of the six faces is removed at random with each dot equally likely to the chosen. If the die is then rolled, then find the probability that the odd number appears.

Explanation

`E_(2):` event that the dot is removed from and odd face:
`P(E_(1))=(1+3+5)/(21)=9/21`
`E_(2):` dot is removed from the even face,
`P(E_(2))=(2+4+6)/(21)=12/21`
E : odd number appears when die is thrown
Also when dot is removed from and odd numbered face, there are exactly 2 faces with odd number of dots, and when dot is removed from an even numbered face, there are exactly 4 faces with odd number of dots.
`therefore` Required probability
`P(E)=P(EnnE_(1))+P(EnnE_(2))`
`=P(E_(1)).P(E//E_(1))+P(E_(2)).P(E//E_(2))`
`=((9)/(21))xx2/9+((12)/(21))xx4/6=11/21`