Let the probability of 2 appearing in a throw be given by P(2) and the probability of 2 not appearing be given by P(2′).
⇒P(2)=61,P(2′)=65
Now, we want 2 to appear in 2ndthrow or4ththrow or6ththrow...and so on
So, required probability is given by,
P(E)=65×61+(65)3×61+(65)5×61+...
⇒P(E)=61×1−362565
⇒P(E)=61×361165
⇒P(E)=65×1136×61
⇒P(E)=115