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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} $

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  • 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)=\(\large\;C^n_r \;P^r\; Q^{n-r}\)
P(at least 8 heads )=P(X=8)=P(X=9)+p(X=10)


answered Feb 27, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1

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