"A)\\;P(X=4)=C_{10}^4(0.3)^4(0.7)^6=0.2001\\\\\nB)\\;P(X\\geq{2})=1-P(X<2)=1-P(X=0)-P(X=1)=1-C_{10}^0(0.3)^0(0.7)^{10}-C_{10}^1(0.3)^1(0.7)^9=0.8507\\\\\nC)\\;P(X>9)=P(X=10)=C_{10}^{10}(0.3)^{10}(0.7)^0=0.000006"
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