Answer
See the picture.
Work Step by Step
$P(x)={}_nC_{x}~p^x~(1-p)^{n-x}$
n = 10, p = 0.5 and 1 - p = 0.5
$P(0)={}_{10}C_{0}\times0.5^0\times0.5^{10}=\frac{10!}{0!\times10!}\times0.5^{10}=1\times0.5^{10}=0.0009766$
$P(1)={}_{10}C_{1}\times0.5^1\times0.5^9=\frac{10!}{1!\times9!}\times0.5^{10}=10\times0.5^{10}=0.009766$
$P(2)={}_{10}C_{2}\times0.5^2\times0.5^8=\frac{10!}{2!\times8!}\times0.5^{10}=45\times0.5^{10}=0.04395$
$P(3)={}_{10}C_{3}\times0.5^3\times0.5^7=\frac{10!}{3!\times7!}\times0.5^{10}=120\times0.5^{10}=0.1172$
$P(4)={}_{10}C_{4}\times0.5^4\times0.5^6=\frac{10!}{4!\times6!}\times0.5^{10}=210\times0.5^{10}=0.2051$
$P(5)={}_{10}C_{5}\times0.5^5\times0.5^5=\frac{10!}{5!\times5!}\times0.5^{10}=252\times0.5^{10}=0.2461$
$P(6)={}_{10}C_{6}\times0.5^6\times0.5^4=\frac{10!}{6!\times4!}\times0.5^{10}=210\times0.5^{10}=0.2051$
$P(7)={}_{10}C_{7}\times0.5^7\times0.5^3=\frac{10!}{7!\times3!}\times0.5^{10}=120\times0.5^{10}=0.1172$
$P(8)={}_{10}C_{8}\times0.5^8\times0.5^2=\frac{10!}{8!\times2!}\times0.5^{10}=45\times0.5^{10}=0.04395$
$P(9)={}_{10}C_{9}\times0.5^9\times0.5^1=\frac{10!}{9!\times1!}\times0.5^{10}=10\times0.5^{10}=0.009766$
$P(10)={}_{10}C_{10}\times0.5^{10}\times0.5^0=\frac{10!}{10!\times0!}\times0.5^{10}=1\times0.5^{10}=0.0009766$
