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# 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}  Can you answer this question? ## 1 Answer 0 votes 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