If ten coins are tossed, then what is the probability of getting atleast 8 heads?

In this case, we have to find out the probability of getting atleast 8 heads. Let X is the random variable for getting a head.

Here, $n = 10$, $r \ge 8$,

i.e., $r = 8,9,10,p = \frac{1}{2},q = \frac{1}{2}$

We know that, $P(X = r){ = ^n}{C_r}{p^r}{q^{n - r}}$

$\therefore$ $P(X = r) = P(r = 8) + P(r = 9) + P(r = 10)$

${ = ^{10}}{C_8}{\left( {\frac{1}{2}} \right)^8}{\left( {\frac{1}{2}} \right)^{10 - 8}}{ + ^{10}}{C_9}{\left( {\frac{1}{2}} \right)^9}{\left( {\frac{1}{2}} \right)^{10 - 9}}{ + ^{10}}{C_{10}}{\left( {\frac{1}{2}} \right)^{10}}{\left( {\frac{1}{2}} \right)^{10 - 10}}$

$= \frac{{10!}}{{8!2!}}{\left( {\frac{1}{2}} \right)^{10}} + \frac{{10!}}{{9!1!}}{\left( {\frac{1}{2}} \right)^{10}} + \frac{{10!}}{{0!10!}}{\left( {\frac{1}{2}} \right)^{10}}$

$= {\left( {\frac{1}{2}} \right)^{10}}\left[ {\frac{{10 \times 9}}{2} + 10 + 1} \right]$

$= {\left( {\frac{1}{2}} \right)^{10}} \cdot 56 = \frac{1}{{{2^7} \cdot {2^3}}} \cdot 56 = \frac{7}{{128}}$

For getting atleast 8 heads, take random variable X for getting head on tossing a coin.

So, get sum of $P(8),P(9)$ and $P(10)$ to get the answer.