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The probability of a man hitting a target is 0.25.He shoots 7 times.What is the probability of his hitting at least twice?

1 Answer

Toolbox:
  • Man hitting a target=0.25
  • p=\(\large\frac{1}{4}\)
  • q=1-\(\large\frac{1}{4}\)=\(\large\frac{3}{4}\)
  • The s 7 timesman shoot
P(hits target atleast 2 times)=P(\(x>2\))
=1-P(\(x<2\))=1-[p(x=0)+(x=1)]
=1-[\(C^7_0\)(\(\large\frac{1}{4}\))\(^0\)(\(\large\frac{3}{4})^7\)+\(C^7_1\)(\(\large\frac{1}{4})\;^1\)(\(\large\frac{3}{6})\;^6\)]
=1-[\(\left(\large\frac{3^7}{4^7}\right)\)+7\(\left(\large\frac{3^6}{4^6}\right)\)]
=\(\Large\frac{9094}{16384}\)
=\(\Large\frac{4547}{8152}\)

 

answered Feb 27, 2013 by poojasapani_1
edited Mar 16, 2013 by poojasapani_1
 

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