If a fair coin is tossed 10 times the probability of atleast 6 heads…

P (atleast 6 heads) = P (6 heads) + P(7 heads) + P (8 heads) + P (9 heads) + P (10 heads)

for n(heads) → Binomial distribution probability

$={ }^{10} C_n \times \frac{1}{2^{10}}$ as each head and tail have $p=\frac{1}{2}$

so for 10 tosses probability $\underbrace{\frac{1}{2} \times \frac{1}{2} ...\frac{1}{2}}_{10 \text { times}}=\frac{1}{2^{10}}$

effective probability = no. of combinations of n heads × $\frac{1}{2^{10}}$

so P (atleast 6 heads) = ${ }^{10} \frac{C_6}{2^{10}}+\frac{{ }^{10} C_7}{2^{10}}+\frac{{ }^{10} C_8}{2^{10}}+{ }^{10} \frac{C_9}{2^{10}}+\frac{{ }^{10} C_{10}}{2^{10}}$

$=\frac{193}{512}$