A fair coin is tossed 10 times. The probability of obtaining exactly six heads is :

$\frac{105}{512}$ $\frac{107}{512}$ $\frac{210}{512}$ $\frac{53}{64}$

$\frac{105}{512}$

The correct answer is Option (1) → $\frac{105}{512}$ P(exactly 6 heads) $={^{10}C}_6×(\frac{1}{2})^6(\frac{1}{2})^4$ $=\frac{10!}{6!(10-6)1}=\frac{1}{2^{10}}$ $=\frac{10×9×8×7}{4×3×2}×\frac{1}{2^{10}}=\frac{105}{512}$