"p(x)=\\frac{e^{-\\lambda}\\lambda^{x}}{x!}, where\\ x=0,1,2,3...\\\\\n\\lambda=5\\\\\np(x>2)=1-p(x\\le2)\\\\\np(x>2)=1-[\\frac{e^{-5}5^{0}}{0!}+\\frac{e^{-5}5^{1}}{1!}+\\frac{e^{-5}5^{2}}{2!}]\\\\\np(x>2)=1-[0.0067+0.0337+0.0842]\\\\\np(x>2)=1-0.1247\\\\\np(x>2)=0.8753"
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