If a fair coin is tossed 10 times, then the probability of obtaining…

P(atleast one head)

= P(one head) + P(two heads) + P(three heads) + ... + P(ten heads)

By Bernoulli's probability distribution

eg: for obtaining (1) head in 10 toses → no of possible ways $={ }^{10} C_1$

for obtaining (2) Heads → possible ways $={ }^{10} C_2$
               .
               .
               .
for 10 heads → possible ways $={ }^{10} C_10$

probability for each case = $\frac{1}{2} \times \frac{1}{2} ... \frac{1}{2}$ (10 times)

$=\frac{1}{2^{10}}$

so, total probability of atleast 1 head

$=\frac{10}{{ }^{10} C_{10}}+\frac{10 C_2}{2^{10}}+...\frac{{ }^{10} C_{10}}{2^{10}}$

P(atleast 1 head) = $\frac{{ }^{10} c_1+{ }^{10} c_2+...{ }^{10} c_{10}}{2^{10}}$

we know that 

$2^n=(1+1)^n={ }^n c_0+{ }^n c_1+{ }^n c_2...{ }^n c_n$

so $2^n=1+{ }^n c_1+{ }^n c_2 ...{ }^n c_n$

$\Rightarrow\left(2^n-1\right)={ }^n c_1+{ }^n c_2 ...{ }^n c_n$

so P(atleast 1 head) = $\frac{2^{10}-1}{2^{10}}=\frac{1024-1}{1024}=\frac{1023}{1024}$