Probability — Class 12 Maths Solution

We know that, $P(X = r){ = ^n}{C_r}{(p)^r}{(q)^{n - r}}$
Here, $n = 10,p = \frac{1}{2},q = \frac{1}{2}$
and $r \ge 8$ i.e., $r = 8,9,10$

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

${ = ^{10}}{C_8}{\left( {\frac{1}{2}} \right)^8}{\left( {\frac{1}{2}} \right)^{10 - 8}}{ + ^{10}}{C_9}{\left( {\frac{1}{2}} \right)^9}\left( {\frac{1}{2}} \right){ + ^{10}}{C_{10}}{\left( {\frac{1}{2}} \right)^{10}} \cdot {\left( {\frac{1}{2}} \right)^0}$

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

$= {\left( {\frac{1}{2}} \right)^{10}} \cdot [45 + 10 + 1] = {\left( {\frac{1}{2}} \right)^{10}} \cdot 56$

$= \frac{1}{{16 \cdot 64}} \cdot 56 = \frac{7}{{128}}$