A) \[\frac{63}{256}\]
B) \[\frac{121}{172}\]
C) \[\frac{113}{512}\]
D) \[\frac{11}{64}\]
Correct Answer: D
Solution :
Probability of getting head,\[p=\frac{1}{2}\]and not getting head,\[q=\frac{1}{2}\] \[\therefore \]Required probability \[{{=}^{10}}{{C}_{7}}{{\left( \frac{1}{2} \right)}^{7}}{{\left( \frac{1}{2} \right)}^{3}}{{+}^{10}}{{C}_{8}}{{\left( \frac{1}{2} \right)}^{8}}{{\left( \frac{1}{2} \right)}^{2}}\] \[{{+}^{10}}{{C}_{9}}{{\left( \frac{1}{2} \right)}^{9}}+{{\left( \frac{1}{2} \right)}^{1}}{{+}^{10}}{{C}_{10}}{{\left( \frac{1}{2} \right)}^{10}}\] \[=120{{\left( \frac{1}{2} \right)}^{10}}+45{{\left( \frac{1}{2} \right)}^{10}}+10{{\left( \frac{1}{2} \right)}^{10}}+1{{\left( \frac{1}{2} \right)}^{10}}\] \[=\frac{176}{1024}=\frac{11}{64}\]
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