Ten coins are tossed.What is the probability of getting at least 8 heads?

\$\begin{array}{1 1}(A)\;\large\frac{7}{128}\$$B)\;\large\frac{8}{122}\\(C)\;\large\frac{128}{7}\\(D)\;\large\frac{5}{128}\end{array}  1 Answer Toolbox: • In a toos of coin getting head and Tail are equally likely • P=P(H)=\(\large\frac{1}{2}$$
• q=1-P=1-$$\large\frac{1}{2}$$=$$\large\frac{1}{2}$$
• Tossed 10 times
• N=10
• X=$$(0\;,1\;,2\;,3\;,4\;,5\;,6\;,7\;,8\;,9\;,10)$$
• P(X=r)=$$\large\;C^n_r \;P^r\; Q^{n-r}$$
$$C^{10}_8\;\Large(\frac{1}{2}\;)^8\;(\frac{1}{2}\;)^{10}+C^{10}_9\;(\frac{1}{2}\;)^9\;(\frac{1}{2}\;)^1+\;C^{10}_{10}(\frac{1}{2}\;)^{10}\;(\frac{1}{2}\;)^0$$
=($$\Large\frac{1}{2}$$)$$^{10}\times$$[45+10+1]
=($$\frac{1}{2}$$)$$^{10}\times$$56
=$$\frac{7}{128}$$

edited Jun 4, 2013