$$6$$ coins are tossed together $$64$$  times. If throwing a head is considered as a success then the expected frequency of at least $$3$$ successes is :

Solution :
$$\eqalign{ & {\text{Here, }}p = \frac{1}{2},\,\,q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2} \cr & n = 6,\,N = 64 \cr & {\text{Then,}} \cr & p\left( r \right) = {}^n{C_r}\,{p^r}{q^{n - r}} \cr & = {}^6{C_r}{\left( {\frac{1}{2}} \right)^r}.{\left( {\frac{1}{2}} \right)^{6 - r}} \cr & = {}^6{C_r}{\left( {\frac{1}{2}} \right)^6} \cr & \therefore \,f\left( r \right) = Np\left( r \right) \cr & = 64.{}^6{C_r}.\frac{1}{{64}} \cr & = {}^6{C_r} \cr & {\text{Now, }}\sum\limits_3^6 {p\left( r \right)} = p\left( 3 \right) + p\left( 4 \right) + p\left( 5 \right) + p\left( 6 \right) \cr & = \left( {{}^6{C_3} + {}^6{C_4} + {}^6{C_5} + {}^6{C_6}} \right)\frac{1}{{{2^6}}} \cr & = \left( {{2^6} - {}^6{C_0} - {}^6{C_1} - {}^6{C_2}} \right)\frac{1}{{{2^6}}} \cr & = \left( {64 - 1 - 6 - 15} \right)\frac{1}{{{2^6}}} \cr & = \frac{{42}}{{64}} \cr & = \frac{{21}}{{32}} \cr & \therefore \,f{\left( r \right)_{r \geqslant 3}} = N\sum\limits_3^6 {p\left( r \right)} = 64.\frac{{21}}{{32}} = 42 \cr} $$