Solution:
Solution
One can get heads after 1 throw or 3 or 5 ... till infinity.
Let probability of a head = p and that of a tail = (1-p).
Hence, probability $$=p+{ (1-p) }^{ 2 }*p+{ (1-p) }^{ 4 }*p+...\infty =\dfrac { p }{ 1-{ (1-p) }^{ 2 } } =\dfrac { 1 }{ 2-p } $$
$$\Rightarrow$$, $$\dfrac { 1 }{ 2-p } =\dfrac { 7 }{ 11 } \quad =>\quad p=\dfrac { 3 }{ 7 } $$
